Differentiation Formula Practice

In this exercise, all functions are assumed to be differentiable and to have all required properties.

For example:

any function in a denominator is assumed to be nonzero
any function inside a logarithm is assumed to be positive
and so on

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

You can ‘slide constants out’ of the differentiation process: $$\frac{d}{dx}\left( kf(x)\right) = k\frac{d}{dx} f(x)$$
The derivative of a sum is the sum of the derivatives: $$\frac{d}{dx} \bigl( f(x) + g(x) \bigr) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$$ (Since subtraction is just adding the opposite, this also works for differences.)
Here, Leibniz notation is used on both sides of the equation.

The equation can also be written using prime notation on the right: $$\frac{d}{dx} \bigl( f(x) + g(x) \bigr) = f'(x) + g'(x)$$
The derivative of a product is NOT the product of the derivatives!
The PRODUCT RULE is: $$\frac{d}{dx}\left( f(x)g(x)\right) = f(x)g'(x) + g(x)f'(x)$$ Verbalize: the derivative of a product is the first, times the derivative of the second, plus the second, times the derivative of the first.

Some people memorize the formula in a different way: $$\frac{d}{dx}\left( f(x)g(x)\right) = f'(x)g(x) + f(x)g'(x)$$ Verbalize: the derivative of a product is the derivative of the first, times the second, plus the first, times the derivative of the second.
The derivative of a quotient is NOT the quotient of the derivatives!
The QUOTIENT RULE is: $$\frac{d}{dx}\left( \frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$$ Verbalize: the derivative of a quotient is the bottom, times the derivative of the top, minus the top, times the derivative of the bottom, all over the bottom squared.

Some people memorize the formula in a different way: $$\frac{d}{dx}\left( \frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$ Verbalize: the derivative of a quotient is the derivative of the top, times the bottom, minus the top, times the derivative of the bottom, all over the bottom squared.
The CHAIN RULE tells how to differentiate composite functions: $$\frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$$ In the expression $\,f(g(x))\,,$ we can call $\,f\,$ the ‘outer function’ and $\,g\,$ the ‘inner function’.
Note also the order of operations indicated by $\,f(g(x))\,$: $\,g\,$ acts on $\,x\,$ first; then $\,f\,$ acts on $\,g(x)\,.$

Verbalize: the derivative of a composite function is the outer function's derivative, evaluated at the inner function, times the derivative of the inner function.

There are two factors in the expression $\,f'(g(x))g'(x)\,$: the first factor is $\,f'(g(x))\,,$ and the second factor is $\,g'(x)\,.$
The expression $\,f'(g(x))\,$ is read aloud as ‘$f$ prime of $g$ of $x$’.

The chain rule is used to extend each ‘basic rule’ in the chart below to the ‘general rule’.
Notice the pattern: replace every $\,x\,$ in the ‘basic rule’ with $\,f(x)\,,$ and then multiply by $\,f'(x)\,.$

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