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: $\arcsin(x) = \sin^{-1}(x)$ $\arccos(x) = \cos^{-1}(x)$ $\arctan(x) = \tan^{-1}(x)$ |
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
IN PROGRESS |