Derivatives Cheat Sheet

Basic Differentiation Rules

Function Form	Derivative
Power Rule: $y = c x^n + C$	$y' = c n x^{n-1}$
Sum/Difference Rule: $y = u(x) \pm v(x)$	$y' = u'(x) \pm v'(x)$
Product Rule: $y = u(x) \cdot v(x)$	$y' = u'(x) v(x) + u(x) v'(x)$
Quotient Rule: $y = \frac{u(x)}{v(x)}$	$y' = \frac{u'(x) v(x) - u(x) v'(x)}{v^2(x)}$
General Power Rule: $y = u(x)^{v(x)}$	$y' = u^v \left( \frac{u' v}{u} + v' \ln u \right)$

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Special Functions & Their Derivatives

Exponential Functions

Function	Derivative
$y = e^x$	$y' = e^x$
$y = e^{kx}$	$y' = k e^{kx}$
$y = a^x$	$y' = a^x \ln a$
$y = e^{u(x)}$	$y' = u'(x) e^{u(x)}$ (Chain Rule)

Logarithmic Functions

Function	Derivative
$y = \ln x$	$y' = \frac{1}{x}$
$y = \log_a x$	$y' = \frac{1}{x \ln a}$
$y = \ln u(x)$	$y' = \frac{u'(x)}{u(x)}$ (Logarithmic Differentiation)

Trigonometric Functions

Function	Derivative
$y = \sin x$	$y' = \cos x$
$y = \cos x$	$y' = -\sin x$
$y = \tan x$	$y' = \sec^2 x = 1 + \tan^2 x$
$y = \cot x$	$y' = -\csc^2 x = -(1 + \cot^2 x)$
$y = \sec x$	$y' = \sec x \tan x$
$y = \csc x$	$y' = -\csc x \cot x$

Inverse Trigonometric Functions

Function	Derivative
$y = \arcsin x$	$y' = \frac{1}{\sqrt{1 - x^2}}$
$y = \arccos x$	$y' = -\frac{1}{\sqrt{1 - x^2}}$
$y = \arctan x$	$y' = \frac{1}{1 + x^2}$
$y = \operatorname{arccot} x$	$y' = -\frac{1}{1 + x^2}$

Hyperbolic Functions

Function	Derivative
$y = \sinh x$	$y' = \cosh x$
$y = \cosh x$	$y' = \sinh x$
$y = \tanh x$	$y' = \text{sech}^2 x$
$y = \coth x$	$y' = -\text{csch}^2 x$

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Advanced Differentiation Techniques

Chain Rule (Derivative of a Composite Function)

If $y = f(u(x))$, then:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(u) \cdot u'(x)$$

Parametric Differentiation

Given $x = f(t), y = g(t)$:

$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{g'(t)}{f'(t)}$$

Second derivative:

$$\frac{d^2y}{dx^2} = \frac{\dot{x} \ddot{y} - \dot{y} \ddot{x}}{\dot{x}^3}$$

Implicit Differentiation

For equations where $y$ is not explicitly expressed in terms of $x$, differentiate both sides and solve for $y'$.

Example:
For $x^2 + y^2 = r^2$,

$$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$$

Derivative of Inverse Functions

If $y = f(x)$ has an inverse $x = f^{-1}(y)$, then:

$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$$

Example:
For $y = \arccos x$,

$$x = \cos y \implies \frac{dx}{dy} = -\sin y \implies \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$$