This web exercise gives practice applying differentiation rules to a variety of differentiation problems.
All the calculus steps are shown in the solutions, but only minimal algebraic simplication is done.

If you want practice with the formulas only, then study Differentiation Formula Practice.

There are two versions of this exercise—you may want to look at both to see which works best for you.

The page you are on uses strict parentheses: for example, $\ln(x)$ is never abbreviated as $\ln x$.
Parentheses are always used to hold the function input.
Also, parentheses are always used for the $\,\frac{d}{dx}\,$ operator: for example, $\,\frac{d}{dx}\bigl(f(x)\bigr)\,$ is never abbreviated as $\,\frac{d}{dx} f(x)\,.$
This often requires more parentheses than might otherwise be used, but eliminates any possible ambiguity in order of operations.
This other version is more casual in its use of parentheses.
For example, when using multi-letter function names (like ‘$\,\sin\,$’ or ‘$\,\ln\,$’),
parentheses are dropped if there is no confusion about the desired order of operations—hence $\,\sin(x)\,$ might be abbreviated as $\,\sin x\,.$
The two versions of this exercise use different verbalizations for the product rule and the quotient rule.

THE DIFFERENTIATION RULES

You can ‘slide constants out’ of the differentiation process: $$\frac{d}{dx} \bigl(kf(x)\bigl) = k\frac{d}{dx}\bigl(f(x)\bigr)$$
The derivative of a sum is the sum of the derivatives: $$\frac{d}{dx} \bigl( f(x) + g(x) \bigr) = \frac{d}{dx} \bigl(f(x)\bigr) + \frac{d}{dx}\bigl(g(x)\bigr)$$ (Since subtraction is just adding the opposite, this also works for differences.)
The derivative of a product is NOT the product of the derivatives!
The PRODUCT RULE is: $$\frac{d}{dx} \bigl(f(x)g(x)\bigr) = f'(x)g(x) + f(x)g'(x)$$ Verbalize: the derivative of a product is the derivative of the first, times the second; plus the first, times the derivative of the second.

Since both multiplication and addition are commutative, there are many different ways to state the product rule.
The version given here has two main advantages:
- the pattern generalizes to products having more than two factors: e.g., $$\bigl(abcd\bigr)' = a'bcd + ab'cd + abc'd + abcd'$$
- if a student encounters some derivative formulas in later math courses where multiplication is not commutative (and hence order matters), they will already have memorized the correct version
The derivative of a quotient is NOT the quotient of the derivatives!
The QUOTIENT RULE is: $$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$ 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: $$\frac{d}{dx}\bigl(f(g(x))\bigr) = f'(g(x))g'(x)$$ The chain rule is used to extend each ‘basic rule’ in the chart below to a more 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	a more general rule	important things to remember
power functions	the Simple Power Rule: $\displaystyle\frac{d}{dx}\bigl(x^n\bigr) = nx^{n-1}$	the General Power Rule: $\displaystyle\frac{d}{dx} \bigl((f(x))^n\bigr) = n\bigl(f(x)\bigr)^{n-1}f'(x)$	power functions have a variable base, and a constant in the exponent; the Power Rules tell how to differentiate power functions
exponential function, base $\text{e}$	$\displaystyle\frac{d}{dx}\bigl({\text{e}}^x\bigr) = {\text{e}}^x$	$\displaystyle\frac{d}{dx}\bigl({\text{e}}^{f(x)}\bigr) = {\text{e}}^{f(x)}\cdot f'(x)$	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 that point
exponential functions, base $a \gt 0$, $a\ne 1$	$\displaystyle\frac{d}{dx}\bigl(a^x\bigr) = a^x\ln(a)$	$\displaystyle\frac{d}{dx} \bigl(a^{f(x)}\bigr) = a^{f(x)}\ln(a)\cdot f'(x)$	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 $\text{e}$	$\displaystyle\frac{d}{dx}\bigl(\ln(x)\bigr) = \frac{1}{x}$	$\displaystyle\frac{d}{dx} \bigl(\ln (f(x))\bigr) = \frac{1}{f(x)}\cdot f'(x)$	the derivatives of logarithmic functions involve the reciprocal of the input
logarithmic functions, base $a \gt 0$, $a\ne 1$	$\displaystyle\frac{d}{dx}\bigl(\log_a(x)\bigr) = \frac{1}{x\ln (a)}$	$\displaystyle\frac{d}{dx}\bigl(\log_a (f(x))\bigr) = \frac{1}{f(x)\ln (a)}\cdot f'(x)$	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	$\displaystyle\frac{d}{dx} \bigl(\sin (x)\bigr) = \cos (x)$ $\displaystyle\frac{d}{dx} \bigl(\cos (x)\bigr) = -\sin (x)$	$\displaystyle\frac{d}{dx} \bigl(\sin (f(x))\bigr) = \cos (f(x))\cdot f'(x)$ $\displaystyle\frac{d}{dx} \bigl(\cos (f(x))\bigr) = -\sin (f(x))\cdot f'(x)$	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	$\displaystyle\frac{d}{dx} \bigl(\tan (x)\bigr) = \sec^2 (x)$ $\displaystyle\frac{d}{dx} \bigl(\cot (x)\bigr) = -\csc^2 (x)$	$\displaystyle\frac{d}{dx} \bigl(\tan (f(x))\bigr) = \sec^2 (f(x))\cdot f'(x)$ $\displaystyle\frac{d}{dx} \bigl(\cot (f(x))\bigr) = -\csc^2 (f(x))\cdot f'(x)$	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	$\displaystyle\frac{d}{dx} \bigl(\sec (x)\bigr) = \sec (x)\tan (x)$ $\displaystyle\frac{d}{dx} \bigl(\csc (x)\bigr) = -\csc (x)\cot (x)$	$\displaystyle\frac{d}{dx} \bigl(\sec (f(x))\bigr) = \sec (f(x))\tan (f(x))\cdot f'(x)$ $\displaystyle\frac{d}{dx} \bigl(\csc f(x)\bigr) = -\csc (f(x))\cot (f(x))\cdot f'(x)$	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	$\displaystyle\frac{d}{dx} \bigl(\arcsin (x)\bigr) = \frac{1}{\sqrt{1-x^2}}$ $\displaystyle\frac{d}{dx} \bigl(\arccos (x)\bigr) = \frac{-1}{\sqrt{1-x^2}}$ $\displaystyle\frac{d}{dx} \bigl(\arctan (x)\bigr) = \frac{1}{1+x^2}$	$\displaystyle\frac{d}{dx} \bigl(\arcsin (f(x))\bigr) = \frac{1}{\sqrt{1-(f(x))^2}}\cdot f'(x)$ $\displaystyle\frac{d}{dx} \bigl(\arccos (f(x))\bigr) = \frac{-1}{\sqrt{1-(f(x))^2}}\cdot f'(x)$ $\displaystyle\frac{d}{dx} \bigl(\arctan (f(x))\bigr) = \frac{1}{1+(f(x))^2}\cdot f'(x)$	alternate notation: $\arcsin(x) = \sin^{-1}(x)$ $\arccos(x) = \cos^{-1}(x)$ $\arctan(x) = \tan^{-1}(x)$
variable base to variable power	Example (the ‘log trick’): Differentiate: $y = x^{2x}$ Rename: $\displaystyle y = x^{2x} = {\text{e}}^{\ln (x^{2x})} = {\text{e}}^{2x\ln (x)}$ (continued in next column)	Differentiate: $\displaystyle \frac{dy}{dx} = {\text{e}}^{2x\ln (x)}\bigl(2\ln (x) + 2x\cdot\frac{1}{x}\bigr)$ Rename back and simplify: $\displaystyle\frac{dy}{dx} = x^{2x}(2\ln (x) + 2)$	Because of the inverse relationship between $\ln (x)$ and ${\text{e}}^x\,,$ we have: $${\text{e}}^{\ln (x)} = x\ \ \text{for all } x \gt 0$$ Also, a property of logarithms is: $$\ln (x^y) = y\ln (x)$$

In the exercises below, the calculus steps in finding the derivatives are shown.
However, only minimal simplification is done:

constants are often multiplied together and pulled to the front
rational exponents are written as radicals

You should be aware that your teacher may desire a different ‘name’ for the answer than the one given here!

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 Integration Practice