Derivative Rules

Definition of the Derivative

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

Basic Rules

• Power Rule: $\frac{d}{dx}[x^n] = nx^{n-1}$
• Constant Multiple: $\frac{d}{dx}[cf] = cf'$
• Sum/Difference: $\frac{d}{dx}[f \pm g] = f' \pm g'$

Product Rule

$\frac{d}{dx}[fg] = f'g + fg'$

Quotient Rule

$\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}$

Chain Rule

$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Trigonometric Derivatives

$\frac{d}{dx}[\sin x] = \cos x, \quad \frac{d}{dx}[\cos x] = -\sin x, \quad \frac{d}{dx}[\tan x] = \sec^2 x$

Exponential and Logarithmic

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

Inverse Trig Derivatives

$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$

Implicit Differentiation

Differentiate both sides with respect to $x$, using the chain rule for $y$ terms (multiply by $dy/dx$).

Problem: Find $\frac{dy}{dx}$ for $y = e^{3x}\sin(2x)$.

Solution: Product rule + chain rule: $y' = 3e^{3x}\sin(2x) + e^{3x} \cdot 2\cos(2x) = e^{3x}[3\sin(2x) + 2\cos(2x)]$

• Know all derivative rules cold: power, product, quotient, chain.
• Derivatives of $e^x$, $\ln x$, trig, and inverse trig are essential.
• Implicit differentiation: chain rule on $y$ terms → solve for $dy/dx$.