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 $u = 2x$:
$\dfrac{dy}{dx} = \dfrac{2}{1 + (2x)^2}$
$\boxed{\dfrac{dy}{dx} = \dfrac{2}{1 + 4x^2}}$
