MIXED DIFFERENTIATION PRACTICE (strict parentheses version)
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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 DIFFERENTIATION RULES

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:   $y = x^{2x}$
  1. Rename: $\displaystyle y = x^{2x} = {\text{e}}^{\ln (x^{2x})} = {\text{e}}^{2x\ln (x)}$
    (continued in next column)
  1. Differentiate: $\displaystyle \frac{dy}{dx} = {\text{e}}^{2x\ln (x)}\bigl(2\ln (x) + 2x\cdot\frac{1}{x}\bigr)$

  2. 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:

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