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⬅ Back to Differential Calculus Topics

Limits and Continuity

Limits describe the behavior of a function as the input approaches a value. Continuity means the limit equals the function value at that point.

$$\lim_{x \to a} f(x) = L$$
$$\text{Continuous at } x=a \iff \begin{cases} f(a) \text{ is defined} \\ \lim_{x \to a} f(x) \text{ exists} \\ \lim_{x \to a} f(x) = f(a) \end{cases} $$

Types of discontinuities: removable, jump, infinite.

Definition of the Derivative

The derivative measures instantaneous rate of change and slope of the tangent line.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
$$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$

Tangent and Normal Lines

1. Equation of the tangent line at $x=a$:

$$y - f(a) = f'(a)(x - a)$$

2. Equation of the normal line at $x=a$:

$$y - f(a) = -\frac{1}{f'(a)}(x - a)$$

Linear Approximation and Differentials — Key Formulas

$$L(x) = f(a) + f'(a)(x-a)$$
$$dy = f'(x)\, dx$$
$$\Delta y \approx dy$$

Evaluating a Limit by Factoring — CE Board

Evaluate: $\displaystyle\lim_{x\to 3}\frac{x^2 - 9}{x - 3}$

Direct substitution gives $0/0$. Factor the numerator:

$\lim_{x\to 3}\dfrac{(x-3)(x+3)}{x-3} = \lim_{x\to 3}(x+3) = 3 + 3 = \boxed{6}$

Derivative from Definition — CE Board

Using the limit definition $f'(x) = \displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$, find $f'(x)$ if $f(x) = x^2$.

$f'(x) = \lim_{h\to 0}\dfrac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}\dfrac{2xh + h^2}{h} = \lim_{h\to 0}(2x + h)$
$\boxed{f'(x) = 2x}$

Equation of Tangent Line — CE Board

Find the equation of the tangent line to $y = x^2$ at the point $(2,\,4)$.

Slope: $y' = 2x$, so at $x = 2$: $m = 4$.

$y - 4 = 4(x - 2)$
$\boxed{y = 4x - 4}$

Equation of Normal Line — CE Board

Find the equation of the normal line to $y = x^3$ at the point $(1,\,1)$.

Slope of tangent: $y' = 3x^2 = 3$ at $x = 1$.

Slope of normal (negative reciprocal): $m_n = -\dfrac{1}{3}$.

$y - 1 = -\dfrac{1}{3}(x - 1)$
$3y - 3 = -(x - 1) \implies \boxed{x + 3y = 4}$

Fundamental Trigonometric Limit — CE Board

Evaluate: $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$

This is the standard trigonometric limit, proved via the squeeze theorem or L'Hôpital's Rule:

$\lim_{x\to 0}\dfrac{\sin x}{x} = \boxed{1}$

This result is used extensively in derivative proofs for $\sin x$.

Corollary: $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x} = 1$ and $\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0$

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q243

MSTE - Differential Calculus / Limits / Engr. Janclyde Espinosa (Clidez)

Evaluate:

q243

Answer:

  1. π
  2. 2π
  3. 3π
  4. 4π

Solution pending in psadquestions/q243.json.

Question Bank: q432

MSTE - Differential Calculus / Angle Between Two Curves / Engr. Janclyde Espinosa (Clidez)

Find the angle of intersection between the curves:
y=x2+1 and y=x+x-1.

Answer:

  1. 63.44º
  2. 71.57º
  3. 45º
  4. 18.43º
Solve the intersection:
$x^2+1=x+x^{-1}$ gives $x=1$, so the point is $(1,2)$. Slopes:
$m_1=2x=2$
$m_2=1-x^{-2}=0$
$\tan\theta=\left|\frac{2-0}{1+0}\right|=2$
$\boxed{\theta=63.44^\circ}$

Question Bank: q433

MSTE - Differential Calculus / Equation of Tangent Line and Normal Line / Engr. Janclyde Espinosa (Clidez)

At what point of the parabola y=x2-3x-5 is the tangent line parallel to 3x-y=3?

Answer:

  1. (3,-5)
  2. (3,5)
  3. (-3,-5)
  4. (-3,5)
The line $3x-y=3$ has slope 3. For $y=x^2-3x-5$:
$y'=2x-3$
$2x-3=3$, so $x=3$. Then:
$y=9-9-5=-5$
$\boxed{(3,-5)}$

Question Bank: q434

MSTE - Differential Calculus / Equation of Tangent Line and Normal Line / Engr. Janclyde Espinosa (Clidez)

For the curve y=x2+x, at what point does the normal line at (0,0) intersect the tangent line at (1,2).

Answer:

  1. (1/4,1/4)
  2. (-1/4,1/4)
  3. (-1/4,-1/4)
  4. (1/4,-1/4)

Solution pending in psadquestions/q434.json.

Question Bank: q458

MSTE - Differential Calculus / Derivatives / Engr. Janclyde Espinosa (Clidez)

Find the angle that the line 2y-9x-18=0 makes with the x-axis.

Answer:

  1. 77.47º
  2. 47.77º
  3. 74.77º
  4. 4.5º
Rewrite the line:
$2y-9x-18=0\Rightarrow y=\frac{9}{2}x+9$
The angle with the x-axis is:
$\theta=\tan^{-1}(9/2)$
$\boxed{77.47^\circ}$

Question Bank: q459

MSTE - Differential Calculus / Angle Between Two Curves / Engr. Janclyde Espinosa (Clidez)

Find the acute angle of intersection between the curves x2=8y and xy=8.

Answer:

  1. 71º34'
  2. 74º31'
  3. 73º14'
  4. 73º41'
The curves intersect where $x^2=8y$ and $xy=8$. Substituting $y=x^2/8$ gives $x=4$, $y=2$. Slopes:
For $x^2=8y$, $y'=x/4=1$. For $xy=8$, $y'=-y/x=-1/2$.
$\tan\theta=\left|\frac{1-(-1/2)}{1+1(-1/2)}\right|=3$
$\boxed{71^\circ34'}$

Question Bank: q460

MSTE - Differential Calculus / Equation of Tangent Line and Normal Line / Engr. Janclyde Espinosa (Clidez)

Find the equation of the tangent to the curve y=x+2x1/3 through point (8,12).

Answer:

  1. 7x-6y+16=0
  2. 8x+5y+21=0
  3. 5x-6y-15=0
  4. 6x-7y+16=0
For $y=x+2x^{1/3}$:
$y'=1+\frac{2}{3}x^{-2/3}$
At $x=8$, $y'=1+\frac{2}{3}(1/4)=7/6$. The tangent through $(8,12)$ is:
$y-12=\frac{7}{6}(x-8)$
$\boxed{7x-6y+16=0}$

Question Bank: t767

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the following limit: $\lim_{x\to0} \frac{5x^2 + 3\cos x - 3}{2x}$.

  1. 0
  2. infinity
  3. 2
  4. 1
At $x = 0$ the form is $\frac{0}{0}$, so apply L'Hôpital's rule:
$\lim_{x\to0} \frac{10x - 3\sin x}{2}$
At $x = 0$: $\frac{0 - 0}{2}$
$\boxed{0}$

Question Bank: t768

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Determine the limit of $(x - 5)/(x^2 - x - 20)$ as $x$ approaches 5.

  1. 0.333
  2. 0.247
  3. 0.111
  4. 0.254
Factor the denominator:
$\frac{x - 5}{x^2 - x - 20} = \frac{x - 5}{(x - 5)(x + 4)} = \frac{1}{x + 4}$
At $x = 5$:
$\frac{1}{9}$
$\boxed{0.111}$

Question Bank: t769

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The value of the $\lim_{x\to0} \frac{x + \tan x}{\sin 3x}$ is:

  1. 0.667
  2. 0.333
  3. 0.543
  4. 0.467
Near $x = 0$, $\tan x \approx x$ and $\sin 3x \approx 3x$:
$\lim_{x\to0} \frac{x + \tan x}{\sin 3x} = \frac{x + x}{3x} = \frac{2}{3}$
$\boxed{0.667}$

Question Bank: t770

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Solve the following limits:

$\lim_{x\to2} \frac{8x - 2x^3}{x^2 - 2x}$.

  1. 16
  2. -6
  3. -8
  4. 12

$\lim_{x\to0} \frac{\tan 2x}{\tan x}$.

  1. 6
  2. 2
  3. 8
  4. 4

$\lim_{x\to\infty} \frac{(2x+1)^3}{x^3} $.

  1. 8
  2. 10
  3. 7
  4. 9

Part 1.

Factor: $\frac{8x - 2x^3}{x^2 - 2x} = \frac{2x(2-x)(2+x)}{x(x-2)} = \frac{-2(2+x)(x-2)}{x-2} = -2(2 + x)$.
At $x = 2$:
$-2(4)$
$\boxed{-8}$

Part 2.

Near $x = 0$, $\tan 2x \approx 2x$ and $\tan x \approx x$:
$\lim_{x\to0} \frac{\tan 2x}{\tan x} = \frac{2x}{x}$
$\boxed{2}$

Part 3.

As $x \to \infty$, only the leading terms matter:
$\frac{(2x+1)^3}{x^3} \to \frac{(2x)^3}{x^3} = \frac{8x^3}{x^3}$
$\boxed{8}$

Question Bank: t773

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Differentiate the following with respect to $x$:

$y = \sin (\sqrt{x} - 1)$.

  1. $\frac{\cos(\sqrt{x}- 1)}{2\sqrt{x}}$
  2. $\frac{\cos(\sqrt{x}- 1)}{2x}$
  3. $\frac{\cos(\sqrt{x}- 1)}{\sqrt{x}}$
  4. $\frac{\cos(\sqrt{x}- 1)}{x}$

$y = \csc^3 (1 - 2x)$.

  1. $6 \csc^2 (1 - 2x) \cot (1 - 2x)$
  2. $6x \csc^2 (1 - 2x) \cot (1 - 2x)$
  3. $6x \csc^3 (1 - 2x) \cot (1 - 2x)$
  4. $6 \csc^3 (1 - 2x) \cot (1 - 2x)$

$y = \left\ A. $1/(1 - x^2)$ 11 Find $dy/dx$ if $y = \ln \sqrt{\frac{x+1}{x- 1}}$.

  1. $1/(1 - x^2)$
  2. $1/(1 + x^2)$
  3. $1/(x^2 - 1)$
  4. $1/(x + 1)$

Part 1.

Chain rule with $u = \sqrt{x} - 1$, $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$:
$\frac{dy}{dx} = \cos(\sqrt{x} - 1)\cdot\frac{1}{2\sqrt{x}}$
$\boxed{\frac{\cos(\sqrt{x} - 1)}{2\sqrt{x}}}$

Part 2.

With $y = \csc^3(1 - 2x)$, apply the chain rule:
$\frac{dy}{dx} = 3\csc^2(1-2x)\cdot\big[-\csc(1-2x)\cot(1-2x)\big]\cdot(-2)$
$= 6\csc^3(1-2x)\cot(1-2x)$
$\boxed{6\csc^3(1 - 2x)\cot(1 - 2x)}$

Question Bank: t776

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

If $f(x) = x(2^x)$, then $f'(x)$ is equal to:

  1. $2^x (\ln 2 + 1)$
  2. $2^x (x \ln 2 + 1)$
  3. $2^x (\ln 2x + 2)$
  4. $2^x (\ln 2x + 1)$
Product rule on $f(x) = x\cdot 2^x$ (recall $\frac{d}{dx}2^x = 2^x\ln 2$):
$f'(x) = (1)(2^x) + x(2^x\ln 2)$
$= 2^x(1 + x\ln 2)$
$\boxed{2^x(x\ln 2 + 1)}$

Question Bank: t778

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Given the parametric equation: $x = 1 + \sin 2\theta$, $y = 1 + \cos \theta + \cos 2\theta$. Find the equation of the tangent at $\theta = 60^\circ$.

  1. $y = - 2.598x + 3.848$
  2. $y = 2.598x - 3.848$
  3. $y = 1.563x - 4.527$
  4. $y = -1.563x + 4.527$
Differentiate parametrically: $\frac{dx}{d\theta} = 2\cos 2\theta$, $\frac{dy}{d\theta} = -\sin\theta - 2\sin 2\theta$.
At $\theta = 60^\circ$: $\frac{dx}{d\theta} = 2\cos 120^\circ = -1$, $\frac{dy}{d\theta} = -\sin 60^\circ - 2\sin 120^\circ = -2.598$.
Slope $= \frac{-2.598}{-1} = 2.598$.
Point: $x = 1 + \sin 120^\circ = 1.866$, $y = 1 + \cos 60^\circ + \cos 120^\circ = 1$.
Tangent: $y - 1 = 2.598(x - 1.866)$
$\boxed{y = 2.598x - 3.848}$

Question Bank: t779

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The parametric equation of the curve is $x = 2\cos^3 \theta$, $y = 2\sin^3 \theta$. Find the equation of the normal to the curve at $\theta = \pi/4$.

  1. $y - x = 0$
  2. $y + x = 0$
  3. $y - 2x = 0$
  4. $y + 2x = 0$
Parametric slope: $\frac{dy}{dx} = \frac{6\sin^2\theta\cos\theta}{-6\cos^2\theta\sin\theta} = -\tan\theta$.
At $\theta = \frac{\pi}{4}$: tangent slope $= -1$, so the normal slope is $+1$.
Point: $x = 2\cos^3 45^\circ = 0.707$, $y = 2\sin^3 45^\circ = 0.707$.
Normal: $y - 0.707 = 1(x - 0.707)$
$\boxed{y - x = 0}$

Question Bank: t784

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Answer the following problems:

Find the equation of the line parallel to 2x + y = 3 and normal to the curve y = x⁴.

  1. 8x + 4y + 9 = 0
  2. 8x - 4y + 9 = 0
  3. 32x + 16y - 17 = 0
  4. 16x - 32y - 17 = 0

Find the angle of intersection of the curves x + xy = 1 and y^3 = (x + 1)^2.

  1. 63.8°
  2. 68.4°
  3. 70.4°
  4. 75.6°

Find the value of the second derivative of the curve y = x^2 e^x at x = -1.

  1. - 0.368
  2. -0.421
  3. -0.587
  4. -0.699

Part 1.

The line $2x + y = 3$ has slope $-2$. The required line is normal to $y = x^4$ with slope $-2$, so the curve's tangent slope there is $\frac{1}{2}$.
$y' = 4x^3 = \frac{1}{2} \Rightarrow x = \frac{1}{2}$, $y = \frac{1}{16}$.
Line of slope $-2$ through $\left(\frac{1}{2}, \frac{1}{16}\right)$:
$y - \frac{1}{16} = -2\left(x - \frac{1}{2}\right)$
$\boxed{32x + 16y - 17 = 0}$

Part 2.

The curves intersect near $(0.44, 1.275)$. Implicit slopes:
$x + xy = 1 \Rightarrow y' = -\frac{1+y}{x} = -5.17$
$y^3 = (x+1)^2 \Rightarrow y' = \frac{2(x+1)}{3y^2} = 0.59$
$\tan\theta = \left|\frac{-5.17 - 0.59}{1 + (-5.17)(0.59)}\right| = 2.81$
$\theta = \tan^{-1}(2.81)$
$\boxed{70.4^\circ}$

Part 3.

$y = x^2 e^x$.
$y' = e^x(x^2 + 2x)$
$y'' = e^x(x^2 + 2x) + e^x(2x + 2) = e^x(x^2 + 4x + 2)$
At $x = -1$: $e^{-1}(1 - 4 + 2) = -e^{-1}$
$\boxed{-0.368}$

Question Bank: t800

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A body moves according to the equation $x(t) = \frac{1- t^2}{3+t}$. What is the velocity of the particle when t = 1?

  1. 1/2
  2. -1/2
  3. 1/3
  4. -1/3
Velocity is $x'(t)$. Using the quotient rule on $x = \frac{1 - t^2}{3 + t}$:
$x'(t) = \frac{(-2t)(3+t) - (1 - t^2)(1)}{(3+t)^2} = \frac{-t^2 - 6t - 1}{(3+t)^2}$
At $t = 1$: $\frac{-1 - 6 - 1}{16} = \frac{-8}{16}$
$\boxed{-\frac{1}{2}}$

Question Bank: t801

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The motion of a particle moving along a straight line is given by the equation $S = 4t^3 - t^4$. What is the maximum displacement of the particle during the interval $-1 \le t \le 4$.

  1. 16
  2. 27
  3. 36
  4. 42
$S = 4t^3 - t^4$: $S' = 12t^2 - 4t^3 = 4t^2(3 - t) = 0 \Rightarrow t = 0, 3$.
Evaluate $S$ at the critical points and endpoints $[-1, 4]$:
$S(-1) = -5,\ S(0) = 0,\ S(3) = 108 - 81 = 27,\ S(4) = 0$
The maximum displacement is:
$\boxed{27}$

Question Bank: t802

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A body moves along a straight line according to the equation $x(t) = 4t^5 - 20t^4$. At what value of "t" when the particle reverses its motion?

  1. 1
  2. 3
  3. 4
  4. 2
The particle reverses where velocity changes sign.
$v = x'(t) = 20t^4 - 80t^3 = 20t^3(t - 4)$
Setting $v = 0$ gives $t = 0$ and $t = 4$. At $t = 4$ the velocity changes from negative to positive, so the motion reverses there.
$\boxed{t = 4}$

Question Bank: t803

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The motion of a body along a straight line is defined by the following equation: $x(t) = t^3 - 8t^2 - 12t - 5$. At what range of "t" is the distance increasing?

  1. -2/3 < t > 6
  2. 6 > t < -2/3
  3. t > 6 only
  4. 6 < t or t < -2/3
Distance increases where velocity is positive.
$v = x'(t) = 3t^2 - 16t - 12$
Roots: $t = \frac{16 \pm \sqrt{256 + 144}}{6} = \frac{16 \pm 20}{6} = 6,\ -\frac{2}{3}$
The upward parabola is positive outside its roots:
$\boxed{6 < t \text{ or } t < -\tfrac{2}{3}}$

Question Bank: t839

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate $\int_0^1 e^x \cos x dx$.

  1. 2.117
  2. 1.324
  3. 2.756
  4. 3.254

Solution pending in psadquestions/t839.json.

Question Bank: t840

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the integral: $\int_{\pi/3}^{2\pi/3} \csc x \cot x dx$.

  1. 1
  2. 0
  3. 1/2
  4. -1
The antiderivative of $\csc x\cot x$ is $-\csc x$:
$\left[-\csc x\right]_{\pi/3}^{2\pi/3} = -\csc\frac{2\pi}{3} + \csc\frac{\pi}{3}$
Since $\sin\frac{2\pi}{3} = \sin\frac{\pi}{3}$, the two terms are equal:
$-1.1547 + 1.1547$
$\boxed{0}$

Question Bank: t841

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the following integrals:

$\int_0^a \frac{x^2 dx}{(a^2+x^2)^2}$.

  1. $\frac{\pi-2}{8a}$
  2. $\frac{\pi-2}{a}$
  3. $\frac{\pi-2}{2a}$
  4. $\frac{\pi-2}{4a}$

$\int_0^{\ln 2} x e^{-x} dx$.

  1. 0.3658
  2. 0.4448
  3. 0.2365
  4. 0.1534

$\int_0^{\pi/2} \cos^8 x dx$.

  1. 0.2458
  2. 0.4295
  3. 0.3254
  4. 0.1225

Part 1.

Substitute $x = a\tan\theta$ ($dx = a\sec^2\theta\,d\theta$); the integrand simplifies to $\frac{\sin^2\theta}{a}$ over $0$ to $\frac{\pi}{4}$:
$\frac{1}{a}\int_0^{\pi/4}\sin^2\theta\,d\theta = \frac{1}{2a}\left[\theta - \frac{\sin 2\theta}{2}\right]_0^{\pi/4} = \frac{1}{2a}\left(\frac{\pi}{4} - \frac{1}{2}\right)$
$\boxed{\frac{\pi - 2}{8a}}$

Part 2.

Integrate by parts: $\int x e^{-x}\,dx = -(x + 1)e^{-x}$.
$\left[-(x+1)e^{-x}\right]_0^{\ln 2} = -(\ln 2 + 1)(0.5) + 1$
$= -0.8466 + 1$
$\boxed{0.1534}$

Part 3.

Use the Wallis formula for $\int_0^{\pi/2}\cos^8 x\,dx$:
$\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{105}{384}\cdot\frac{\pi}{2}$
$\boxed{0.4295}$

Question Bank: t844

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the following integrals:

$\int_0^1 \frac{x^2 dx}{(x+1)^4}$.

  1. 0.0672
  2. 0.0821
  3. 0.0417
  4. 0.0234

$\int_0^1 \arcsin y dy$.

  1. 0.772
  2. 0.335
  3. 0.571
  4. 0.234

$\int_0^{\pi/2} (\sin x)^2 (\cos x)^6 dx$.

  1. 0.0998
  2. 0.0123
  3. 0.0432
  4. 0.0614

Part 1.

Substitute $u = x + 1$ ($x = u - 1$, $x^2 = u^2 - 2u + 1$):
$\int_1^2 \frac{u^2 - 2u + 1}{u^4}\,du = \left[-\frac{1}{u} + \frac{1}{u^2} - \frac{1}{3u^3}\right]_1^2$
$= -0.2917 - (-0.3333)$
$\boxed{0.0417}$

Part 2.

Integrate by parts:
$\int_0^1 \arcsin y\,dy = \left[y\arcsin y + \sqrt{1 - y^2}\right]_0^1$
$= \left(1\cdot\frac{\pi}{2} + 0\right) - (0 + 1) = \frac{\pi}{2} - 1$
$\boxed{0.571}$

Part 3.

Use the Wallis formula for $\int_0^{\pi/2}\sin^2 x\cos^6 x\,dx$ (both even):
$\frac{(1)(5\cdot3\cdot1)}{8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2} = \frac{15}{384}\cdot\frac{\pi}{2}$
$\boxed{0.0614}$

Question Bank: t847

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate $\int_0^{\pi} e^x \sin x dx$.

  1. 2.16
  2. 2.05
  3. 1.82
  4. 1.94

Solution pending in psadquestions/t847.json.

Question Bank: t848

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the following: $\int_2^3 x\tan(x) dx$.

  1. - 2.3524
  2. -1.8236
  3. -0.5696
  4. -2.015
Integrate by parts ($\int\tan x\,dx = -\ln|\cos x|$):
$\int x\tan x\,dx = -x\ln|\cos x| + \int \ln|\cos x|\,dx$
The remaining integral has no elementary form, so evaluate numerically over $[2, 3]$:
$\boxed{\approx -2.015}$

Question Bank: t849

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate the following integrals:

$\int_3^6 \frac{x^3}{x-4} dx$.

  1. 209.36
  2. 234.56
  3. 298.43
  4. 267.43

(Reconstructed) Evaluate the integral involving $\text{arcsec}(x)$.

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)

(Reconstructed) Evaluate $\int_1^4 \cos(2/x) dx$.

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)
Polynomial division: $\frac{x^3}{x-4} = x^2 + 4x + 16 + \frac{64}{x-4}$.
$\int_3^6 = \left[\frac{x^3}{3} + 2x^2 + 16x + 64\ln|x-4|\right]_3^6$
$= (72 + 72 + 96 + 64\ln 2) - (9 + 18 + 48 + 0)$
$= 284.36 - 75$
$\boxed{209.36}$

Question Bank: t853

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

(Reconstructed) The motion of a particle is defined by $v_x = 2t^2$, $v_y = 4 - t^2$, and $v_z = 8t$.

What is the velocity of the particle when $t = 4$ seconds?

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)

(Reconstructed) What is the acceleration of the particle when $t = 4$ seconds?

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)

(Reconstructed) What is the distance travelled by the particle from $t = 0$ to $t = 4$ seconds?

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)

Solution pending in psadquestions/t853.json.

Question Bank: t856

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

(Reconstructed) A body moves such that its distance is $S = 12t^3 - 2t^2 + 20$. Determine its acceleration when $t = 4$.

  1. (Option A)
  2. (Option B)
  3. (Option C)
  4. (Option D)

Solution pending in psadquestions/t856.json.

Question Bank: t909

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Find the general solution of the following differential equation: y" + 3y' - 4y = 0.

  1. y = C₁ e⁴ˣ + C₂ xe⁻ˣ
  2. y = C₁ e⁻⁴ˣ + C₂ xeˣ
  3. y = C₁ e⁴ˣ + C₂ e⁻ˣ
  4. y = C₁ e⁻⁴ˣ + C₂ eˣ
Form the characteristic equation:
$r^2 + 3r - 4 = 0 \Rightarrow (r + 4)(r - 1) = 0$
Roots $r = -4$ and $r = 1$ give:
$\boxed{y = C_1 e^{-4x} + C_2 e^{x}}$

Question Bank: t911

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The amount of germs in a certain culture increases at a rate proportional to the number present. If the amount of germs doubles every year, how many years will the number of germs triple?

  1. 1.952
  2. 4.214
  3. 2.568
  4. 1.585
Exponential growth $N = N_0 e^{kt}$. Doubling every year gives $k = \ln 2$.
For tripling, $3 = e^{kt} \Rightarrow t = \frac{\ln 3}{\ln 2}$
$\boxed{t = 1.585 \text{ years}}$

Question Bank: t1004

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

Evaluate $\lim_{x\to0} \left[ \frac{\tan 2x - 2\sin x}{x^3} \right]$.

  1. 0
  2. undefined
  3. 3
  4. infinity
Using Taylor series ($x \to 0$):
$\tan 2x = 2x + \frac{(2x)^3}{3} + \cdots = 2x + \frac{8x^3}{3} + \cdots$
$2\sin x = 2x - \frac{x^3}{3} + \cdots$
$\tan 2x - 2\sin x = \frac{8x^3}{3} + \frac{x^3}{3} + \cdots = 3x^3 + \cdots$
$\lim_{x \to 0} \frac{\tan 2x - 2\sin x}{x^3} = \frac{3x^3}{x^3}$
$\boxed{= 3}$

Question Bank: t1373

MSTE - Differential Calculus / Limits / BEMz

Evaluate the lim $(x^{2}-16)/(x-4)$.

  1. 1
  2. 8
  3. 0
  4. 16

Solution pending in psadquestions/t1373.json.

Question Bank: t1374

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit $(x-4)/(x^{2}-x-12)$ as x approaches 4.

  1. undefined
  2. 0
  3. infinity
  4. 1/7

Solution pending in psadquestions/t1374.json.

Question Bank: t1375

MSTE - Differential Calculus / Limits / BEMz

What is the limit of cos (1/y) as y approaches infinity?

  1. 0
  2. -1
  3. infinity
  4. 1

Solution pending in psadquestions/t1375.json.

Question Bank: t1376

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limits of lim $(x^{3}-2x+9) /(2x^{3}-8)$.

  1. 0
  2. -9/8
  3. $\infty$
  4. ½

Solution pending in psadquestions/t1376.json.

Question Bank: t1377

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(x^{3}-2x^{2}-x+2) /(x^{2}-4)$ as x approaches 2.

  1. $\infty$
  2. ¾
  3. 2/5
  4. 4/7

Solution pending in psadquestions/t1377.json.

Question Bank: t1378

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $\sqrt{x - 4}/√(x^{2}$ -16) as x approaches 4.

  1. 0.262
  2. 0.354
  3. 0
  4. $\infty$

Solution pending in psadquestions/t1378.json.

Question Bank: t1379

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(x^{2}-x-6)/(x^{2}-4x+3)$ as x approaches 3.

  1. 3/2
  2. 3/5
  3. 0
  4. 5/2

Solution pending in psadquestions/t1379.json.

Question Bank: t1380

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(4x^{2}- x)/ (2x^{2}+4)$ as x approaches ∞.

  1. 2
  2. 4
  3. $\infty$
  4. 0

Solution pending in psadquestions/t1380.json.

Question Bank: t1381

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(x-2)/(x^{3}-8)$ as x approaches 2.

  1. $\infty$
  2. 1/12
  3. 0
  4. 2/3

Solution pending in psadquestions/t1381.json.

Question Bank: t1382

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of θ/(2 sinθ) as θ approaches 0.

  1. 2
  2. ½
  3. 0
  4. $\infty$

Solution pending in psadquestions/t1382.json.

Question Bank: t1383

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(1-\sec^{2}$ (x)/ cos (x)-1 as x approaches 0.

  1. -2
  2. $\infty$
  3. 0
  4. 1

Solution pending in psadquestions/t1383.json.

Question Bank: t1384

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit $(x^{3}-27)/(x-3)$ as x approaches to 3.

  1. 0
  2. infinity
  3. 9
  4. 27

Solution pending in psadquestions/t1384.json.

Question Bank: t1385

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit $(3x^{3}-4x^{2}-5x+2)/ (x^{2}-x-2)$ as x approaches to 2.

  1. $\infty$
  2. 5
  3. 0
  4. 7/3

Solution pending in psadquestions/t1385.json.

Question Bank: t1386

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of (4 $\tan^{3}$ (x)/ 2sin(x)-x as x approaches 0.

  1. 1
  2. 0
  3. 2
  4. $\infty$

Solution pending in psadquestions/t1386.json.

Question Bank: t1387

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of 8x/(2x- 1) as x approaches ∞.

  1. 4
  2. 3
  3. 2
  4. -1

Solution pending in psadquestions/t1387.json.

Question Bank: t1388

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(x^{2}-1)/ (x^{2}+3x-4)$ as x approaches 1.

  1. 2/5
  2. 1/5
  3. 3/5
  4. 4/5

Solution pending in psadquestions/t1388.json.

Question Bank: t1389

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of (x+2)/(x- 2) as x approaches ∞.

  1. $\infty$
  2. -1
  3. 1
  4. 4

Solution pending in psadquestions/t1389.json.

Question Bank: t1390

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit of $(1-cosx)/(x^{2})$ as x approaches 0.

  1. $\infty$
  2. ½
  3. 1
  4. 0

Solution pending in psadquestions/t1390.json.

Question Bank: t1391

MSTE - Differential Calculus / Limits / BEMz

Find the limit of $[\sqrt{x+4}-2]/x$ as x approaches 0.

  1. $\infty$
  2. ¼
  3. 0
  4. ½

Solution pending in psadquestions/t1391.json.

Question Bank: t1392

MSTE - Differential Calculus / Limits / BEMz

Find the limit $[\sqrt{x+9}-3]/x$ as x approaches 0.

  1. $\infty$
  2. 1/6
  3. 0
  4. 1/3

Solution pending in psadquestions/t1392.json.

Question Bank: t1393

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit $(x^{2}+x-6)/(x^{2}-4)$ as x approaches to 2.

  1. 6/5
  2. 5/4
  3. 4/3
  4. 3/2

Solution pending in psadquestions/t1393.json.

Question Bank: t1394

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit $(x^{4}-81)/(x-3)$ as x approaches to 3.

  1. 108
  2. 110
  3. 122
  4. 100

Solution pending in psadquestions/t1394.json.

Question Bank: t1395

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit (x+sin2x)/ (x-sin2x) as x approaches to 0.

  1. -5
  2. -3
  3. 4
  4. -1

Solution pending in psadquestions/t1395.json.

Question Bank: t1396

MSTE - Differential Calculus / Limits / BEMz

Evaluate the limit (ln sin x)/(ln tan x) as x approaches to 0.

  1. 1
  2. 2
  3. ½
  4. $\infty$

Solution pending in psadquestions/t1396.json.

Question Bank: t1397

MSTE - Differential Calculus / Limits / BEMz

Compute the equation of the vertical asymptote of the curve $y=(2x-1)/(x+2)$.

  1. $x+2=0$
  2. $x-3=0$
  3. $x+3=0$
  4. $x-2=0$

Solution pending in psadquestions/t1397.json.

Question Bank: t1398

MSTE - Differential Calculus / Limits / BEMz

Compute the equation of the horizontal asymptote of the curve $y=(2x-1)/(x+2)$.

  1. $y=2$
  2. $y=0$
  3. $y-1=0$
  4. $y-3=0$

Solution pending in psadquestions/t1398.json.

Question Bank: t1399

MSTE - Differential Calculus / Limits / BEMz

The function $y=(x-4)/(x+2)$ is discontinuous at x equals?

  1. -2
  2. 0
  3. 1
  4. 2

Solution pending in psadquestions/t1399.json.

Question Bank: t1414

MSTE - Differential Calculus / Derivatives / BEMz

Find y’ if $y=\ln$ x

  1. 1/x
  2. $\ln x^{2}$
  3. $1/\ln x$
  4. $x \ln x$

Solution pending in psadquestions/t1414.json.

Question Bank: t1415

MSTE - Differential Calculus / Derivatives / BEMz

Find y’ if $y=arc$ sin (x)

  1. $\sqrt{1 -x^{2}}$
  2. $1/\sqrt{1 -x^{2}}$
  3. $1/(1+x^{2})$
  4. $(1+x)/\sqrt{ 1-x^{2}}$

Solution pending in psadquestions/t1415.json.

Question Bank: t1421

MSTE - Differential Calculus / Derivatives / BEMz

If ln (ln y) + ln y = ln x, find y’.

  1. x/(x+y)
  2. x/(x-y)
  3. y/(x+y)
  4. y/(x-y)

Solution pending in psadquestions/t1421.json.

Question Bank: t1422

MSTE - Differential Calculus / Derivatives / BEMz

Find y” if $y=a^{u}$.

  1. $a^{u} \ln a$
  2. $u \ln a$
  3. $a^{u}/\ln a$
  4. $a \ln u$

Solution pending in psadquestions/t1422.json.

Question Bank: t1424

MSTE - Differential Calculus / Derivatives / BEMz

If $y=\tanh$ x, find dy/dx.

  1. $sech^{2} (x)$
  2. $csch^{2} (x)$
  3. $\sinh^{2} (x)$
  4. $\tanh^{2} (x)$

Solution pending in psadquestions/t1424.json.

Question Bank: t1432

MSTE - Differential Calculus / Derivatives / BEMz

If $y1=2x+4$ and $y2=x^{2}+C$, find the value of C such that y2 is tangent to y1.

  1. 6
  2. 5
  3. 7
  4. 4

Solution pending in psadquestions/t1432.json.

Question Bank: t1435

MSTE - Differential Calculus / Derivatives / BEMz

Find y’ in the following equation $y=4x^{2}-3x-1$.

  1. 8x-3
  2. 4x-3
  3. 2x-3
  4. 8x-x

Solution pending in psadquestions/t1435.json.

Question Bank: t1436

MSTE - Differential Calculus / Derivatives / BEMz

Differentiate the equation $y=(x^{2})/(x+1)$.

  1. $(x^{2}+2x)/(x+1)^{2}$
  2. x/(x+1)
  3. $2x^{2}/(x+1)$
  4. 1

Solution pending in psadquestions/t1436.json.

Question Bank: t1437

MSTE - Differential Calculus / Derivatives / BEMz

If $y=x/(x+1)$, find y’.

  1. $1/(x+1)^{3}$
  2. $1/(x+1)^{2}$
  3. x+1
  4. $(x+1)^{2}$

Solution pending in psadquestions/t1437.json.

Question Bank: t1438

MSTE - Differential Calculus / Derivatives / BEMz

Find dy/dx in the equation $y=(x^{6}+3x^{2}+50)/(x^{2}+1)$ if $x=1$

  1. -21
  2. -18
  3. 10
  4. 16

Solution pending in psadquestions/t1438.json.

Question Bank: t1446

MSTE - Differential Calculus / Derivatives / BEMz

Find the equation of the tangent to the curve $y=2e^{x}$ at (0,2).

  1. $2x-y+3=0$
  2. $2x-y+2=0$
  3. $3x+y+2=0$
  4. $2x+3y+2=0$

Solution pending in psadquestions/t1446.json.

Question Bank: t1464

MSTE - Differential Calculus / Derivatives / BEMz

The equation $y^{2}=cx$ is the general equation of.

  1. $y’=2y/x$
  2. $y’=2x/y$
  3. $y’=y/2x$
  4. $y’=x/2y$

Solution pending in psadquestions/t1464.json.

Question Bank: t1614

MSTE - Differential Calculus / Parametric Equations / BEMz

A particle moves in a plane according to the parametric equations of motions: $x=t^{2}, y=t^{3}$. Find the magnitude of the acceleration when $t=2/3$.

  1. 4.47
  2. 5.10
  3. 4.90
  4. 6.12
Acceleration is the second derivative of position.
$$x=t^2,\quad y=t^3$$
$$\frac{d^2x}{dt^2}=2,\quad \frac{d^2y}{dt^2}=6t$$
At $t=2/3$:
$$\vec a=(2,4)$$
The magnitude is
$$|\vec a|=\sqrt{2^2+4^2}=\sqrt{20}=4.47$$
$$\boxed{4.47}$$

Question Bank: t1619

MSTE - Differential Calculus / Parametric Equations / BEMz

A particle moves along a path whose parametric equations are $x=t^{3}$ and $y=2t^{2}$. What is the acceleration when $t=3\sec$.

  1. $18.44 m/\sec^{2}$
  2. $15.93 m/\sec^{2}$
  3. $23.36 m/\sec^{2}$
  4. $10.59 m/\sec^{2}$
Acceleration is obtained by differentiating each coordinate twice.
$$x=t^3,\qquad y=2t^2$$
$$\frac{d^2x}{dt^2}=6t,\qquad \frac{d^2y}{dt^2}=4$$
At $t=3$:
$$\vec a=(18,4)$$
The acceleration magnitude is
$$|\vec a|=\sqrt{18^2+4^2}=\sqrt{340}=18.44$$
$$\boxed{18.44\text{ m/sec}^2}$$

Question Bank: t2073

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

Evaluate $\frac{dy}{dx}$ when $x = 1$ if $y = 3e^{4x} - \frac{5}{2e^{3x}} + 8\ln 5x$.

  1. 634.5
  2. 678.4
  3. 628.3
  4. 663.6
  5. 685.2
Differentiate term by term:
$y=3e^{4x}-\frac{5}{2e^{3x}}+8\ln5x=3e^{4x}-\frac{5}{2}e^{-3x}+8\ln5x$
$\frac{dy}{dx}=12e^{4x}+\frac{15}{2}e^{-3x}+\frac{8}{x}$
At $x=1$:
$\frac{dy}{dx}=12e^4+\frac{15}{2}e^{-3}+8$
$\boxed{\frac{dy}{dx}\approx663.6}$

Question Bank: t2081

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

Given $f(\theta) = 5\ln 2\theta - 4\ln 3\theta$. Determine $f'(\theta)$.

  1. $\frac{1}{2}\theta$
  2. $\frac{3}{4}\theta$
  3. $\theta$
  4. $2\theta$
  5. $1/\theta$
Use $\frac{d}{d\theta}\ln(k\theta)=\frac{1}{\theta}$ for constant $k$.
$f(\theta)=5\ln2\theta-4\ln3\theta$
$f'(\theta)=5\left(\frac{1}{\theta}\right)-4\left(\frac{1}{\theta}\right)$
$\boxed{f'(\theta)=\frac{1}{\theta}}$

Question Bank: t2098

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

If $y = \frac{3}{x^2} - 2\sin 4x + \frac{2}{e^x} + \ln 5x$, determine $\frac{dy}{dx}$.

  1. $-8\cos 4x - \frac{6}{x^3} + \frac{2}{e^x} + \frac{1}{2x}$
  2. $-10\cos 4x - \frac{6}{x^3} + \frac{2}{e^x} + \frac{2}{x}$
  3. $-6\cos 4x - \frac{6}{x^3} + \frac{2}{e^x} + \frac{1}{x^2}$
  4. $-8\cos 4x - \frac{6}{x^3} - \frac{2}{e^x} + \frac{1}{x}$
  5. $-8\cos 4x - \frac{4}{x^3} - \frac{2}{e^x} - \frac{1}{x}$
Differentiate term by term:
$\frac{d}{dx}\left(\frac{3}{x^2}\right)=\frac{d}{dx}(3x^{-2})=-6x^{-3}=-\frac{6}{x^3}$
$\frac{d}{dx}(-2\sin4x)=-8\cos4x$
$\frac{d}{dx}\left(\frac{2}{e^x}\right)=\frac{d}{dx}(2e^{-x})=-2e^{-x}=-\frac{2}{e^x}$
$\frac{d}{dx}(\ln5x)=\frac{1}{x}$
$\boxed{-8\cos4x-\frac{6}{x^3}-\frac{2}{e^x}+\frac{1}{x}}$

Question Bank: t2124

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

If $f(x) = 4x^5 - 2x^3 + x - 3$, find $y''(x)$.

  1. $60x^3 - 12x$
  2. $80x^3 - 15x$
  3. $80x^3 - 10x$
  4. $80x^2 - 12x$
  5. $80x^3 - 12x$
Differentiate twice.
$f(x)=4x^5-2x^3+x-3$
$f'(x)=20x^4-6x^2+1$
$f''(x)=80x^3-12x$
$\boxed{80x^3-12x}$

Question Bank: t2127

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

If $y = 3x^4 + 2x^3 - 3x + 2$, find $\frac{d^2 y}{dx^2}$.

  1. $36x^2 + 10x$
  2. $36x^2 + 13x$
  3. $36x^2 + 12x$
  4. $36x^2 + 15x$
  5. $36x^2 + 11x$
Differentiate twice.
$y=3x^4+2x^3-3x+2$
$y'=12x^3+6x^2-3$
$y''=36x^2+12x$
$\boxed{36x^2+12x}$

Question Bank: t2130

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

Differentiate with respect to $x$ if $y = 5e^{3x}$.

  1. $15e^{3x}$
  2. $13e^{3x}$
  3. $14e^{3x}$
  4. $12e^{3x}$
  5. $10e^{3x}$
Use $\frac{d}{dx}(e^{u})=e^u\frac{du}{dx}$.
$y=5e^{3x}$
$\frac{dy}{dx}=5e^{3x}(3)$
$\boxed{\frac{dy}{dx}=15e^{3x}}$

Question Bank: t2139

MSTE - Differential Calculus / Limits / Besavilla CE Pre-Board Math & Surveying

Evaluate the following limit: $\lim_{x \to 0} \frac{\tan x - x}{x - \sin x}$

  1. 1
  2. 3
  3. 2
  4. 0
  5. 4
Use the standard series near $x=0$.
$\tan x=x+\frac{x^3}{3}+\cdots$ and $\sin x=x-\frac{x^3}{6}+\cdots$
$\tan x-x=\frac{x^3}{3}+\cdots$
$x-\sin x=\frac{x^3}{6}+\cdots$
$\lim_{x\to0}\frac{\tan x-x}{x-\sin x}=\frac{1/3}{1/6}$
$\boxed{2}$

Question Bank: t2161

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

Differentiate with respect to $x$ if $y = \frac{2}{7e^{2x}}$.

  1. $\frac{-6}{7e^{2x}}$
  2. $\frac{-2}{7e^{2x}}$
  3. $\frac{-5}{7e^{2x}}$
  4. $\frac{-3}{7e^{2x}}$
  5. $\frac{-4}{7e^{2x}}$
Rewrite the function as $y=\frac{2}{7}e^{-2x}$.
$\frac{dy}{dx}=\frac{2}{7}(-2)e^{-2x}$
$\frac{dy}{dx}=-\frac{4}{7e^{2x}}$
$\boxed{\frac{dy}{dx}=\frac{-4}{7e^{2x}}}$

Question Bank: t2163

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

Differentiate the following equation $f(t) = \frac{4}{3e^{5t}}$.

  1. $\frac{-20}{3e^{5t}}$
  2. $\frac{-22}{3e^{5t}}$
  3. $\frac{-25}{3e^{5t}}$
  4. $\frac{-30}{3e^{5t}}$
  5. $\frac{-28}{3e^{5t}}$
Rewrite the function using a negative exponent.
$f(t)=\frac{4}{3e^{5t}}=\frac{4}{3}e^{-5t}$
$f'(t)=\frac{4}{3}(-5)e^{-5t}$
$f'(t)=-\frac{20}{3e^{5t}}$
$\boxed{f'(t)=\frac{-20}{3e^{5t}}}$

Question Bank: t2166

MSTE - Differential Calculus / Derivatives / Besavilla CE Pre-Board Math & Surveying

If $f(t) = 4\ln t + 2$, evaluate $f'(t)$ when $t = 0.25$.

  1. 16
  2. 15
  3. 14
  4. 17
  5. 18
Differentiate $f(t)=4\ln t+2$.
$f'(t)=\frac{4}{t}$
At $t=0.25$: $f'(0.25)=\frac{4}{0.25}$
$\boxed{f'(0.25)=16}$
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