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Math III

Differentiation techniques

Combine the basic derivative formulas with the product, quotient and chain rules to differentiate a wide range of functions.

Basic, Standard: Grade 12 · Term 1 / Advanced: Grade 12 · Exam prep

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Derivatives of basic functions

Formulas
(xα)′=αxα−1,(sin⁡x)′=cos⁡x,(cos⁡x)′=−sin⁡x,(tan⁡x)′=1cos⁡2x(x^{\alpha})' = \alpha x^{\alpha - 1},\quad (\sin x)' = \cos x,\quad (\cos x)' = -\sin x,\quad (\tan x)' = \frac{1}{\cos^2 x}
(ex)′=ex,(ax)′=axlog⁡a,(log⁡x)′=1x,(log⁡∣x∣)′=1x,(log⁡ax)′=1xlog⁡a(e^x)' = e^x,\quad (a^x)' = a^x\log a,\quad (\log x)' = \frac{1}{x},\quad (\log|x|)' = \frac{1}{x},\quad (\log_a x)' = \frac{1}{x\log a}

Here log⁡\log denotes the natural logarithm (base ee).

Product, quotient and chain rules

Rules
{f(x)g(x)}′=f′(x)g(x)+f(x)g′(x),{f(x)g(x)}′=f′(x)g(x)−f(x)g′(x){g(x)}2\{f(x)g(x)\}' = f'(x)g(x) + f(x)g'(x),\qquad \left\{\frac{f(x)}{g(x)}\right\}' = \frac{f'(x)g(x) - f(x)g'(x)}{\{g(x)\}^2}
{f(g(x))}′=f′(g(x)) g′(x)(dydx=dydu⋅dudx)\{f(g(x))\}' = f'(g(x))\,g'(x)\qquad \left(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\right)

Examples: {(2x+1)5}′=5(2x+1)4⋅2=10(2x+1)4\{(2x + 1)^5\}' = 5(2x + 1)^4 \cdot 2 = 10(2x + 1)^4 and (x2ex)′=2xex+x2ex=(x2+2x)ex(x^2e^x)' = 2xe^x + x^2e^x = (x^2 + 2x)e^x.

Common mistake

Forgetting to multiply by the derivative of the inside function: (sin⁡3x)′=3cos⁡3x(\sin 3x)' = 3\cos 3x.

Logarithmic, implicit and parametric differentiation

Logarithmic differentiation: for y=xxy = x^x, take logs to get log⁡y=xlog⁡x\log y = x\log x, differentiate to get y′y=log⁡x+1\frac{y'}{y} = \log x + 1, so y′=xx(log⁡x+1)y' = x^x(\log x + 1).

Implicit differentiation: for x2+y2=25x^2 + y^2 = 25, treat yy as a function of xx: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.

Parametric curves: if x=f(t)x = f(t) and y=g(t)y = g(t), then dydx=dy/dtdx/dt\displaystyle\frac{dy}{dx} = \frac{dy/dt}{dx/dt} and d2ydx2=ddt(dydx)dx/dt\displaystyle\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}.

Higher derivatives

Differentiating y′y' again gives y′′y''; differentiating nn times gives y(n)y^{(n)}. Compute y′y', y′′y'', y′′′y''' and look for a pattern.

(eax)(n)=aneax,(sin⁡x)(n)=sin⁡(x+nπ2),(1x)(n)=(−1)nn!xn+1(e^{ax})^{(n)} = a^ne^{ax},\qquad (\sin x)^{(n)} = \sin\left(x + \frac{n\pi}{2}\right),\qquad \left(\frac{1}{x}\right)^{(n)} = \frac{(-1)^n n!}{x^{n+1}}