The functions you met earlier — \(\sin x\), \(\cos x\), \(\tan x\), \(e^x\) and \(\ln x\) — each have a standard derivative. When the angle or the argument is itself a function of \(x\), multiply by its derivative: this is the chain rule again.
Trigonometric functions (angles in radians)
| Function | Derivative | Example |
|---|---|---|
| \(y=\sin x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\cos x\) | \(y=4\sin x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4\cos x\) |
| \(y=\cos x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\sin x\) | \(y=5\cos x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-5\sin x\) |
| \(y=\tan x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sec^2 x\) | \(y=\dfrac{1}{3}\tan x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{3}\sec^2 x\) |
Watch the sign of \(\cos x\): its derivative is \(-\sin x\), the opposite sign.
Trigonometric functions with the chain rule
| Function | Derivative | Example |
|---|---|---|
| \(y=\sin f(x)\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\cos f(x)\cdot f'(x)\) | \(y=\sin(4x+1)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4\cos(4x+1)\) |
| \(y=\cos f(x)\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\sin f(x)\cdot f'(x)\) | \(y=\cos\left(x^2\right)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-2x\sin\left(x^2\right)\) |
| \(y=\tan f(x)\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\sec^2 f(x)\cdot f'(x)\) | \(y=\tan 5x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=5\sec^2 5x\) |
| \(y=\sin^n x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=n\sin^{n-1}x\cdot\cos x\) | \(y=\sin^4 x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4\sin^3 x\cos x\) |
Exponential functions
| Function | Derivative | Example |
|---|---|---|
| \(y=e^x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=e^x\) | \(y=7e^x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=7e^x\) |
| \(y=e^{f(x)}\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=e^{f(x)}\cdot f'(x)\) | \(y=e^{x^2+1}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2xe^{x^2+1}\) |
| \(y=e^{ax+b}\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=ae^{ax+b}\) | \(y=4e^{2x-3}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=8e^{2x-3}\) |
Logarithmic functions
| Function | Derivative | Example |
|---|---|---|
| \(y=\ln x\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{x}\) | \(y=2\ln x\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2}{x}\) |
| \(y=\ln f(x)\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{f'(x)}{f(x)}\) | \(y=\ln\left(x^2+9\right)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2x}{x^2+9}\) |
| \(y=\ln(ax+b)\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{a}{ax+b}\) | \(y=\ln(3x+7)\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{3}{3x+7}\) |
Simplify first with the laws of logarithms
| Simplify | Then differentiate |
|---|---|
| \(y=\ln\sqrt{\dfrac{x+2}{3x-1}}\) \(=\dfrac{1}{2}[\ln|x+2|-\ln|3x-1|]\) | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2}\left[\dfrac{1}{x+2}-\dfrac{3}{3x-1}\right]\) |
Work in radians. Numbers such as \(\pi\), \(e^4\) and \(\ln 7\) are constants — they differentiate to \(0\). Apply the laws of logarithms before differentiating whenever the expression allows it.
The original function \(\ln\sqrt{(x+2)/(3x-1)}\) has domain \(x<-2\) or \(x>1/3\). Absolute values in the split logarithms preserve both branches; the derivative formula applies on this original domain. In general, \(\ln u\) requires \(u>0\).
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