Differentiation Formula Practice
DIFFERENTIATION FORMULA PRACTICE
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In this exercise, all functions are assumed to be differentiable and to have all required properties.

For example:

This web exercise gives practice with all the following differentiation rules:

to differentiate: the basic rule the general rule important things to remember
power functions the Simple Power Rule:
$\displaystyle\frac{d}{dx} x^n = nx^{n-1}$
the General Power Rule:
$\displaystyle\frac{d}{dx} (f(x))^n = n\bigl(f(x)\bigr)^{n-1}f'(x)$
power functions have a variable base,
and a constant in the exponent;
the Power Rules tell how to differentiate power functions
exponential function, base $\text{e}$ $\displaystyle\frac{d}{dx} {\text{e}}^x = {\text{e}}^x$ $\displaystyle\frac{d}{dx} {\text{e}}^{f(x)} = {\text{e}}^{f(x)}\cdot f'(x)$ exponential functions have a constant base,
and the variable in the exponent;
for this special function, its derivative is itself—the $y$-value of a point on the graph of $y = {\text{e}}^x$ is the same as the slope of the tangent line at the point
exponential functions, $a \gt 0$, $a\ne 1$ $\displaystyle\frac{d}{dx} a^x = a^x\ln a$ $\displaystyle\frac{d}{dx} a^{f(x)} = a^{f(x)}(\ln a)\cdot f'(x)$ exponential functions have a constant base,
and the variable in the exponent;
note that $\,\ln\text{e} = 1\,$, so the constant disappears when the base is the irrational number $\,\text{e}$
logarithmic function, base $\text{e}$ $\displaystyle\frac{d}{dx} \ln{x} = \frac{1}{x}$ $\displaystyle\frac{d}{dx} \ln f(x) = \frac{1}{f(x)}\cdot f'(x)$ the derivatives of logarithmic functions involve the reciprocal of the input
logarithmic functions,
$a \gt 0$, $a\ne 1$
$\displaystyle\frac{d}{dx} \log_a{x} = \frac{1}{x\ln a}$ $\displaystyle\frac{d}{dx} \log_a f(x) = \frac{1}{f(x)\ln a}\cdot f'(x)$ note that $\,\ln\text{e} = 1\,$, so the constant disappears when the base of the logarithm is the irrational number $\,\text{e}$
trigonometric functions, sine and cosine $\displaystyle\frac{d}{dx} \sin x = \cos x$

$\displaystyle\frac{d}{dx} \cos x = -\sin x$
$\displaystyle\frac{d}{dx} \sin f(x) = (\cos f(x))\cdot f'(x)$

$\displaystyle\frac{d}{dx} \cos f(x) = (-\sin f(x))\cdot f'(x)$
sine and cosine are ‘co-functions’:
note the pattern: to find the derivative of a co-function, replace each function in the derivative formula by its co-function, and introduce a minus sign
trigonometric functions, tangent and cotangent $\displaystyle\frac{d}{dx} \tan x = \sec^2 x$

$\displaystyle\frac{d}{dx} \cot x = -\csc^2 x$
$\displaystyle\frac{d}{dx} \tan f(x) = (\sec^2 f(x))\cdot f'(x)$

$\displaystyle\frac{d}{dx} \cot f(x) = (-\csc^2 f(x))\cdot f'(x)$
tangent and cotangent are ‘co-functions’:
note the pattern: to find the derivative of a co-function, replace each function in the derivative formula by its co-function, and introduce a minus sign
trigonometric functions, secant and cosecant $\displaystyle\frac{d}{dx} \sec x = \sec x\tan x$

$\displaystyle\frac{d}{dx} \csc x = -\csc x\cot x$
$\displaystyle\frac{d}{dx} \sec f(x) = (\sec f(x)\tan f(x))\cdot f'(x)$

$\displaystyle\frac{d}{dx} \csc f(x) = (-\csc f(x)\cot f(x))\cdot f'(x)$
secant and cosecant are ‘co-functions’:
note the pattern: to find the derivative of a co-function, replace each function in the derivative formula by its co-function, and introduce a minus sign
inverse trigonometric functions $\displaystyle\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}$

$\displaystyle\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1-x^2}}$

$\displaystyle\frac{d}{dx} \arctan x = \frac{1}{1+x^2}$
$\displaystyle\frac{d}{dx} \arcsin f(x) = \frac{1}{\sqrt{1-(f(x))^2}}\cdot f'(x)$

$\displaystyle\frac{d}{dx} \arccos f(x) = \frac{-1}{\sqrt{1-(f(x))^2}}\cdot f'(x)$

$\displaystyle\frac{d}{dx} \arctan f(x) = \frac{1}{1+(f(x))^2}\cdot f'(x)$
alternate notation:

$\arcsin(x) = \sin^{-1}(x)$

$\arccos(x) = \cos^{-1}(x)$

$\arctan(x) = \tan^{-1}(x)$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Mixed Derivative Practice

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.