Differentiation techniques

Constants and single terms of $x$

Chain rule, product rule, quotient rule

Chain rule: $$ \boxed{ {d \over dx} [f(x)]^n = n [f(x)]^{n - 1} . \phantom{.} f'(x) } $$

Product rule: If $u$ and $v$ are functions of $x$, $$ \boxed{ {d \over dx} (uv) = u{dv \over dx} + v {du \over dX} } $$

Quotient rule: If $u$ and $v$ are functions of $x$, $$ \boxed{ {d \over dx} \left(u \over v\right) = {v {du \over dx} - u {dv \over dx} \over v^2} } $$

Exponential terms

Besides using the formula for $a^{f(x)}$, you can use implicit differentiation: \begin{align} \text{Let } & y = a^{f(x)} \\ \\ \ln y & = \ln a^{f(x)} \\ \ln y & = f(x) \ln a \phantom{00000} [\text{Power law (logarithms)}] \\ \\ {d \over dx} (\ln y) & = {d \over dx} [ (\ln a) f(x) ] \phantom{00000} [\text{Note: } \ln a \text{ is a constant}] \\ {1 \over y} \left( {dy \over dx} \right) & = (\ln a ) f'(x) \\ {dy \over dx} & = \ln a \phantom{.} . \phantom{.} f'(x) \phantom{.} . \phantom{.} y \\ \\ \text{Since } & y = a^{f(x)}, \\ {dy \over dx} & = \ln a \phantom{.} . \phantom{.} f'(x) \phantom{.} . \phantom{.} a^{f(x)} \end{align}

Logarithms

Besides using the formula for $\log_a f(x) $, you can use the change-of-base formula to change to base e: \begin{align} \text{Let } & y = \log_a f(x) \\ \\ y & = {\ln f(x) \over \ln a} \phantom{000000000} [\text{Change base to base } e] \\ y & = {1 \over \ln a} \ln f(x) \phantom{00000} [\text{Note: } \ln a \text{ is a constant}] \\ {dy \over dx} & = {1 \over \ln a} \left[ f'(x) \over f(x)\right] \end{align}

For both cases, try to simplify the expression by logarithm laws (Power, Product & Quotient) before differentiating. For example, \begin{align} {d \over dx} \left[ \ln \sqrt{x \over x + 1} \right] & = {d \over dx} \left[ {1 \over 2} \ln {x \over x + 1} \right] \\ & = {d \over dx} \left[ {1 \over 2} \ln x - {1 \over 2} \ln (x + 1) \right] \\ & = {1 \over 2} \left(1 \over x\right) - {1 \over 2} \left(1 \over x + 1\right) \\ & = {1 \over 2x} - {1 \over 2(x + 1)} \end{align}

Trigonometric functions

Inverse trigonometric functions

For example, \begin{align} {d \over dx} \left[ \tan^{-1} (2x + 1) \right] & = {2 \over 1 + (2x + 1)^2} \\ & = {2 \over 1 + 4x^2 + 4x + 1} \\ & = {2 \over 4x^2 + 4x + 2} \\ & = {1 \over 2x^2 + 2x + 1} \end{align}

Implicit differentiation

Common derivatives: \begin{align} {d \over dx} (y) & = {dy \over dx} \\ \\ {d \over dx} \left(dy \over dx\right) & = {d^2y \over dx^2} \\ \\ {d \over dx} (e^y) & = {dy \over dx} e^y \\ \\ {d \over dx} (\ln y) & = { {dy \over dx} \over y } \\ \\ {d \over dx} (\sin y) & = {dy \over dx} \cos y \end{align}

Examples of implicit differentiation & chain rule: \begin{align} {d \over dx} (y^2) & = 2y {dy \over dx} \\ \\ {d \over dx} \left[ \left(dy \over dx\right)^3 \right] & = 3 \left(dy \over dx \right)^2 \left(d^2y \over dx^2\right) \end{align}

Examples of implicit differentiation & product rule: \begin{align} {d \over dx} (xy) & = x{dy \over dx} + y(1) \\ & = x{dy \over dx} + y \\ \\ {d \over dx} \left(y {dy \over dx}\right) & = y {d^2y \over dx^2} + \left(dy \over dx\right) \left(dy \over dx\right) \\ & = y {d^2y \over dx^2} + \left(dy \over dx\right)^2 \end{align}