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Basic Differentiation Rules

$$\frac{d}{dx}[c] = 0$$
$$\frac{d}{dx}[x^n] = nx^{n-1}$$
$$\frac{d}{dx}[c f(x)] = c f'(x)$$
$$\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$$

Product, Quotient, and Chain Rules

$$\frac{d}{dx}(uv) = u'v + uv'$$
$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$
$$\frac{d}{dx}[f(g(x))] = f'(g(x))\, g'(x)$$

Derivatives of Trigonometric Functions

$$\frac{d}{dx}(\sin x) = \cos x$$
$$\frac{d}{dx}(\cos x) = -\sin x$$
$$\frac{d}{dx}(\tan x) = \sec^2 x$$
$$\frac{d}{dx}(\cot x) = -\csc^2 x$$
$$\frac{d}{dx}(\sec x) = \sec x \tan x$$
$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

Derivatives of Inverse Trigonometric Functions

$$\frac{d}{dx}(\sin^{-1} u) = \frac{u'}{\sqrt{1 - u^2}}$$
$$\frac{d}{dx}(\cos^{-1} u) = \frac{-u'}{\sqrt{1 - u^2}}$$
$$\frac{d}{dx}(\tan^{-1} u) = \frac{u'}{1 + u^2}$$
$$\frac{d}{dx}(\cot^{-1} u) = \frac{-u'}{1 + u^2}$$
$$\frac{d}{dx}(\sec^{-1} u) = \frac{u'}{|u|\sqrt{u^2 - 1}}$$
$$\frac{d}{dx}(\csc^{-1} u) = \frac{-u'}{|u|\sqrt{u^2 - 1}}$$

Derivatives of Exponential and Logarithmic Functions

$$\frac{d}{dx}[e^x] = e^x$$
$$\frac{d}{dx}[\ln x] = \frac{1}{x}$$
$$\frac{d}{dx}[a^x] = a^x \ln a$$
$$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$$

Derivative of a Polynomial — CE Board

Find $y'$ if $y = 3x^4 - 5x^2 + 2x - 7$.

Apply the power rule term by term:

$y' = 12x^3 - 10x + 2$

Product Rule — CE Board

Find the derivative of $f(x) = x^2 \cos x$.

Product rule: $(uv)' = u'v + uv'$.

$f'(x) = 2x\cos x + x^2(-\sin x)$
$\boxed{f'(x) = 2x\cos x - x^2\sin x}$

Quotient Rule — CE Board

Find $\dfrac{d}{dx}\!\left[\dfrac{x^2+1}{x-1}\right]$.

Quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.

$= \dfrac{(2x)(x-1) - (x^2+1)(1)}{(x-1)^2} = \dfrac{2x^2 - 2x - x^2 - 1}{(x-1)^2}$
$\boxed{= \dfrac{x^2 - 2x - 1}{(x-1)^2}}$

Chain Rule with Trig — CE Board

Find $y'$ if $y = \sin(3x^2)$.

Chain rule: outer function $\sin(u)$, inner $u = 3x^2$.

$y' = \cos(3x^2)\cdot\dfrac{d}{dx}(3x^2) = \cos(3x^2)\cdot 6x$
$\boxed{y' = 6x\cos(3x^2)}$

Slope of Logarithmic Curve — CE Board

Find the slope of the curve $y = x\ln x$ at $x = e$.

Differentiate using the product rule:

$y' = (1)(\ln x) + x\cdot\dfrac{1}{x} = \ln x + 1$

At $x = e$:

$y'\big|_{x=e} = \ln e + 1 = 1 + 1 = \boxed{2}$

Derivative of Inverse Tangent — CE Board

Find $\dfrac{dy}{dx}$ if $y = \arctan(2x)$.

Using the chain rule with $\dfrac{d}{dx}[\arctan u] = \dfrac{u'}{1+u^2}$ where $u = 2x$:

$\dfrac{dy}{dx} = \dfrac{2}{1 + (2x)^2}$
$\boxed{\dfrac{dy}{dx} = \dfrac{2}{1 + 4x^2}}$
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q431

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

At what point is 2 the slope of the curve y=4x+x2?

Answer:

  1. (-1,-3)
  2. (-2,-4)
  3. (2,12)
  4. (0,0)
Differentiate $y=4x+x^2$:
$y'=4+2x$
Set the slope equal to 2:
$4+2x=2$, so $x=-1$. Then:
$y=4(-1)+(-1)^2=-3$
$\boxed{(-1,-3)}$

Question Bank: q456

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

Find the slope of the line tangent to the curve y=x3-2x+1 at x=1.

Answer:

  1. 1
  2. 1/2
  3. 1/3
  4. 1/4
Differentiate:
$y'=3x^2-2$
At $x=1$:
$y'=3(1)^2-2$
$\boxed{1}$

Question Bank: q457

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

Determine the slope of the curve x2+y2-6x-4y-21=0 at (0,7).

Answer:

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

Solution pending in psadquestions/q457.json.

Question Bank: t781

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

Given circle x^2 + y^2 = 5 and ellipse 5x^2 + y^2 = 6.

Which of the following is a point of intersection of the curves?

  1. (2.179, 0.5)
  2. (0.5, 2.179)
  3. (0.8, 1.875)
  4. (1.875, 0.8)

What is the slope of the ellipse at the point of intersection?

  1. ±0.524
  2. ±1.147
  3. ±0.229
  4. ±1.358

Find the acute angle of intersection.

  1. 36°
  2. 32°
  3. 34°
  4. 40°

Part 1.

Subtract the circle from the ellipse: $(5x^2 + y^2) - (x^2 + y^2) = 6 - 5$:
$4x^2 = 1 \Rightarrow x = \pm 0.5$
Then $y^2 = 5 - 0.25 = 4.75 \Rightarrow y = \pm 2.179$
$\boxed{(0.5,\ 2.179)}$

Part 2.

Differentiate the ellipse $5x^2 + y^2 = 6$ implicitly:
$10x + 2yy' = 0 \Rightarrow y' = -\frac{5x}{y}$
At $(0.5, 2.179)$: $y' = -\frac{5(0.5)}{2.179} = -1.147$
$\boxed{\pm 1.147}$

Part 3.

Circle slope at $(0.5, 2.179)$: $y' = -\frac{x}{y} = -0.2295$. Ellipse slope $= -1.147$.
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{-0.2295 + 1.147}{1 + (0.2295)(1.147)}\right| = 0.726$
$\theta = \tan^{-1}(0.726)$
$\boxed{36^\circ}$

Question Bank: t787

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

Determine the angle between the polar vector and the tangent to the curve at the indicated angles:

r = a sec (2θ) at θ = π/8.

  1. 33.65°
  2. 26.56°
  3. 32.21°
  4. 29.65°

r = a sin (θ/2) at θ = π/2.

  1. 61.21°
  2. 63.44°
  3. 65.98°
  4. 64.32°

r = θ^2 at θ = π.

  1. 57.52°
  2. 65.78°
  3. 60.33°
  4. 72.22°

Part 1.

The angle $\psi$ between the radius vector and tangent satisfies $\tan\psi = \frac{r}{dr/d\theta}$.
$r = a\sec 2\theta \Rightarrow \frac{dr}{d\theta} = 2a\sec 2\theta\tan 2\theta$, so $\tan\psi = \frac{1}{2\tan 2\theta}$.
At $\theta = \frac{\pi}{8}$ ($2\theta = 45^\circ$): $\tan\psi = \frac{1}{2}$
$\boxed{26.56^\circ}$

Part 2.

$\tan\psi = \frac{r}{dr/d\theta}$. With $r = a\sin\frac{\theta}{2}$, $\frac{dr}{d\theta} = \frac{a}{2}\cos\frac{\theta}{2}$:
$\tan\psi = \frac{a\sin(\theta/2)}{\frac{a}{2}\cos(\theta/2)} = 2\tan\frac{\theta}{2}$
At $\theta = \frac{\pi}{2}$ ($\theta/2 = 45^\circ$): $\tan\psi = 2$
$\boxed{63.44^\circ}$

Part 3.

$\tan\psi = \frac{r}{dr/d\theta}$. With $r = \theta^2$, $\frac{dr}{d\theta} = 2\theta$:
$\tan\psi = \frac{\theta^2}{2\theta} = \frac{\theta}{2}$
At $\theta = \pi$: $\tan\psi = \frac{\pi}{2} = 1.571$
$\boxed{57.52^\circ}$

Question Bank: t791

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

Determine the slope of the following polar curves at the indicated angle: r^2 = cos 2x at x = π/10.

  1. -0.3256
  2. - 0.5214
  3. -0.7265
  4. -0.6325
For a polar curve, $\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$.
From $r^2 = \cos 2\theta$: $r' = \frac{-\sin 2\theta}{r}$. At $\theta = \frac{\pi}{10}$: $r = \sqrt{\cos 36^\circ} = 0.899$, $r' = -0.654$.
Numerator $= (-0.654)(0.309) + (0.899)(0.951) = 0.654$
Denominator $= (-0.654)(0.951) - (0.899)(0.309) = -0.899$
$\frac{dy}{dx} = \frac{0.654}{-0.899}$
$\boxed{-0.7265}$

Question Bank: t797

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

Given the polar curve $r = 4 \sin^2 \theta$.

What is the slope of the curve at $\theta = \pi/6$?

  1. 2.598
  2. 1.039
  3. 2.874
  4. 1.254

What is the rate of change of the slope with respect to x at $\theta = \pi/6$?

  1. 1.586
  2. 1.112
  3. 1.344
  4. 1.982

What is the radius of curvature of the curve at $\theta = \pi/6$.

  1. 2.23
  2. 13.6
  3. 25.34
  4. 5.64

Part 1.

Polar slope: $\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$. With $r = 4\sin^2\theta$, $r' = 4\sin 2\theta$.
At $\theta = \frac{\pi}{6}$: $r = 1$, $r' = 3.464$.
Numerator $= 3.464(0.5) + 1(0.866) = 2.598$
Denominator $= 3.464(0.866) - 1(0.5) = 2.5$
$\frac{dy}{dx} = \frac{2.598}{2.5}$
$\boxed{1.039}$

Part 2.

The rate of change of the slope with $x$ is $\frac{d^2y}{dx^2} = \dfrac{\frac{d}{d\theta}(dy/dx)}{dx/d\theta}$.
With $r''=8\cos2\theta=4$ at $\theta=\frac{\pi}{6}$, computing $\frac{d}{d\theta}(dy/dx) = \frac{N'D - ND'}{D^2} = \frac{7.5(2.5) - 2.598(-0.866)}{2.5^2} = 3.36$.
Divide by $dx/d\theta = D = 2.5$:
$\frac{3.36}{2.5}$
$\boxed{1.344}$

Part 3.

Radius of curvature:
$R = \frac{(1 + (dy/dx)^2)^{3/2}}{|d^2y/dx^2|} = \frac{(1 + 1.039^2)^{3/2}}{1.344}$
$= \frac{(2.08)^{3/2}}{1.344}$
$\boxed{R = 2.23}$

Question Bank: t822

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

If two numbers differ by 8, then what is the least possible value of their product?

  1. 9
  2. -9
  3. 16
  4. -16
Let the numbers be $x$ and $x - 8$. Their product is:
$P = x(x - 8) = x^2 - 8x$
$P' = 2x - 8 = 0 \Rightarrow x = 4$
$P = 4(4 - 8) = 4(-4)$
$\boxed{-16}$

Question Bank: t852

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

(Reconstructed) The slope of a curve at any point is given by $dy/dx = 8x^3 + 6x - 8$. If the curve passes through the point $(1, -1)$, find the equation of the curve.

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

Solution pending in psadquestions/t852.json.

Question Bank: t864

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

The length of the curve $y = 2x^{3/2} + 1$ between $x = 0$ to $x = 6$ is:

  1. 25.47
  2. 28.36
  3. 32.54
  4. 30.14
Arc length $L = \int_0^6 \sqrt{1 + (y')^2}\,dx$, with $y' = 3x^{1/2}$:
$L = \int_0^6 \sqrt{1 + 9x}\,dx = \left[\frac{(1 + 9x)^{3/2}}{13.5}\right]_0^6$
$= \frac{55^{3/2} - 1}{13.5}$
$\boxed{L = 30.14}$

Question Bank: t1431

MSTE - Differential Calculus / Derivatives / BEMz

Determine the slope of the curve $y=x^{2}-3x$ as it passes through the origin.

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

Solution pending in psadquestions/t1431.json.

Question Bank: t1433

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of $(x^{2})y=8$ at the point (2,2).

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

Solution pending in psadquestions/t1433.json.

Question Bank: t1441

MSTE - Differential Calculus / Derivatives / BEMz

What is the slope of the tangent to $y=(x^{2}+1)(x^{3}-4x)$ at (1,-6)?

  1. -8
  2. -4
  3. 3
  4. 5

Solution pending in psadquestions/t1441.json.

Question Bank: t1443

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the curve $x^{2}+y^{2}-6x+10y+5=0$ at point (1,0).

  1. 2/5
  2. ¼
  3. 2
  4. 2

Solution pending in psadquestions/t1443.json.

Question Bank: t1444

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the ellipse $x^{2}+4y^{2}-10x+16y+5=0$ at the point where $y=2+8^{0}.5$ and $x=7$.

  1. -0.1654
  2. -0.1538
  3. -0.1768
  4. -0.1463

Solution pending in psadquestions/t1444.json.

Question Bank: t1445

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the tangent to the curve $y=2x-x^{2}+x^{3}$ at (0,2).

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

Solution pending in psadquestions/t1445.json.

Question Bank: t1447

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the curve $y=2(1+3x)^{2}$ at point (0,3).

  1. 12
  2. -9
  3. 8
  4. -16

Solution pending in psadquestions/t1447.json.

Question Bank: t1448

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the curve $y=x^{2}(x+2)^{3}$ at point (1,2).

  1. 81
  2. 48
  3. 64
  4. 54

Solution pending in psadquestions/t1448.json.

Question Bank: t1449

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the curve $y=[(4-x)^{2}]/x$ at point (2,2).

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

Solution pending in psadquestions/t1449.json.

Question Bank: t1450

MSTE - Differential Calculus / Derivatives / BEMz

If the slope of the curve $y^{2}=12x$ is equal to 1 at point (x,y), find the value of x and y.

  1. $x=3, y=6$
  2. $x=4, y=5$
  3. $x=2, y=7$
  4. $x=5, y=6$

Solution pending in psadquestions/t1450.json.

Question Bank: t1451

MSTE - Differential Calculus / Derivatives / BEMz

If the slope of the curve $x^{2}+y^{2}=25$ is equal to -3/4 at point (x,y) find the value of x and y.

  1. 3,4
  2. 2,3
  3. 3,4.2
  4. 3.5,4

Solution pending in psadquestions/t1451.json.

Question Bank: t1452

MSTE - Differential Calculus / Derivatives / BEMz

If the slope of the curve $25x^{2}+4y^{2}=100$ is equal to -15/8 at point (x,y), find the value of x and y.

  1. 1.2,4
  2. 2,4
  3. 1.2,3
  4. 2,4.2

Solution pending in psadquestions/t1452.json.

Question Bank: t1453

MSTE - Differential Calculus / Derivatives / BEMz

Determine the point on the curve $x^{3}-9x-y=0$ at which slope is 18.

  1. $x=3, y=0$
  2. $x=4, y=5$
  3. $x=2, y=7$
  4. $x=5, y=6$

Solution pending in psadquestions/t1453.json.

Question Bank: t1465

MSTE - Differential Calculus / Derivatives / BEMz

Find the slope of the curve $y=6(4+x)^{1}/2$ at point (0,12).

  1. 1.5
  2. 2.2
  3. 1.8
  4. 2.8

Solution pending in psadquestions/t1465.json.

Question Bank: t1615

MSTE - Differential Calculus / Parametric Equations / BEMz

A particle moves along the right-hand part of the curve $4y^{3}=x^{2}$ with a speed $Vy=dy/dx=constant$ at 2. Find the speed of motion when $y=4$.

  1. 12.17
  2. 14.10
  3. 15.31
  4. 16.40
The statement appears to mean the vertical component of velocity is constant: $dy/dt=2$. For the right-hand branch,
$$4y^3=x^2$$
$$x=2y^{3/2}$$
Differentiate with respect to y:
$$\frac{dx}{dy}=3\sqrt{y}$$
Then
$$\frac{dx}{dt}=\frac{dx}{dy}\frac{dy}{dt}=3\sqrt{y}(2)$$
At $y=4$:
$$\frac{dx}{dt}=3(2)(2)=12$$
The speed is
$$v=\sqrt{12^2+2^2}=12.17$$
$$\boxed{12.17}$$

Question Bank: t1649

MSTE - Differential Calculus / Derivatives / BEMz

If m is the slope of the tangent line to the curve $y=x^{2}-2x^{2}+x$ at the point (x,y), find the instantaneous rate of change of the slope m per unit change in x at the point (2,2).

  1. 8
  2. 9
  3. 10
  4. 11
The printed curve appears to have a source typo. The keyed answer matches the intended curve $y=x^3-2x^2+x$.
The slope of the tangent is
$$m=\frac{dy}{dx}=3x^2-4x+1$$
The instantaneous rate of change of the slope per unit change in x is
$$\frac{dm}{dx}=6x-4$$
At $x=2$:
$$\frac{dm}{dx}=6(2)-4=8$$
$$\boxed{8}$$

Question Bank: t2145

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

Find the horizontal asymptote of the curve $y = \frac{2x^4}{x^4 - 3x^2 - 1}$.

  1. $y^2 - 2 = 0$
  2. $y + 3 = 0$
  3. $y - 3 = 0$
  4. $y + 2 = 0$
  5. $y - 2 = 0$
For a rational function with the same highest degree in numerator and denominator, the horizontal asymptote is the ratio of the leading coefficients.
$y=\frac{2x^4}{x^4-3x^2-1}$
Leading coefficient ratio $=\frac{2}{1}=2$
Thus the horizontal asymptote is $y=2$, or
$\boxed{y-2=0}$