Calculus Reference

Exponential and Logarithmic Derivatives

When the variable is in an exponent or logarithm, keep the function’s form and multiply by the derivative of its inside expression.

Solved Examples

Start with a basic rule, then use the chain rule for a composite input.

Derivative Formula Table

FunctionDerivativeCondition
$e^{u(x)}$$e^{u(x)}u'(x)$All real $u$.
$a^{u(x)}$$a^{u(x)}\ln(a)u'(x)$$a>0$.
$\ln|u(x)|$$\frac{u'(x)}{u(x)}$$u(x)\ne0$.
$\log_a|u(x)|$$\frac{u'(x)}{u(x)\ln(a)}$$a>0$, $a\ne1$, and $u(x)\ne0$.

The absolute value in the logarithm formulas covers both positive and negative nonzero inputs. For $\ln(u)$ without absolute value, the original function requires $u>0$.

Exponential Functions

The natural exponential $e^u$ keeps its form when differentiated. A different positive base adds a factor of $\ln(a)$. If the exponent is a function, multiply by its derivative.

Natural Base
$$\frac{d}{dx}[e^{u(x)}]=e^{u(x)}u'(x)$$
Any Positive Base
$$\frac{d}{dx}[a^{u(x)}]=a^{u(x)}\ln(a)u'(x)$$

Example: Natural Exponential with a Composite Exponent

Differentiate $f(x)=e^{2x^2-1}$

The outside function is $e^u$ and the inside is $u=2x^2-1$.

Solution Steps
  1. Keep the exponential and multiply by the inside derivative.
  2. Differentiate $2x^2-1$ and simplify.
$$f'(x)=e^{2x^2-1}(4x)$$$$\boxed{f'(x)=4xe^{2x^2-1}}$$

Example: Exponential with a Common Base

Differentiate $g(x)=3^{x^2+1}$

Use the constant-base rule and remember its $\ln(3)$ factor.

Solution Steps
  1. Keep $3^{x^2+1}$ and multiply by $\ln(3)$.
  2. Multiply by the derivative of the exponent, $2x$.
$$g'(x)=3^{x^2+1}\ln(3)(2x)$$$$\boxed{g'(x)=2x\ln(3)\,3^{x^2+1}}$$

Logarithmic Functions

The natural logarithm has derivative $1/u$ with respect to its input. A composite input adds the chain-rule factor $u'$. A logarithm with base $a$ has an additional divisor of $\ln(a)$.

Natural Logarithm
$$\frac{d}{dx}[\ln|u(x)|]=\frac{u'(x)}{u(x)}$$
Base-$a$ Logarithm
$$\frac{d}{dx}[\log_a|u(x)|]=\frac{u'(x)}{u(x)\ln(a)}$$

Example: Natural Logarithm with an Inside Function

Differentiate $h(x)=\ln(3x+2)$

The inside expression is positive when $x>-\frac23$.

Solution Steps
  1. Divide the inside derivative by the inside expression.
  2. Simplify and keep the domain of the original logarithm.
$$h'(x)=\frac{3}{3x+2}$$$$\boxed{h'(x)=\frac{3}{3x+2}},\quad x>-\frac23$$

Example: Logarithm with a Common Base

Differentiate $p(x)=\log_2(x^2+1)$

The quadratic input is positive for every real $x$.

Solution Steps
  1. Use $u'/[u\ln(2)]$ for a base-$2$ logarithm.
  2. Substitute $u=x^2+1$ and $u'=2x$.
$$p'(x)=\frac{2x}{(x^2+1)\ln(2)}$$$$\boxed{p'(x)=\frac{2x}{(x^2+1)\ln(2)}}$$

Continue the Derivative Sequence

Use the chain rule for nested functions, then continue to trigonometric derivatives or inverse-trigonometric derivatives.