MathipediaThe free math encyclopedia

Derivative Tricks

In calculus 1, students are taught about the definition of a derivative, f(x)=f(x+h)f(x)hf'(x)=\frac{f(x+h)-f(x)}{h}, which is used to generically define the derivative of a function ff. This definition is useful because it is able to differentiate any function ff at any point xx, but in certain situations also simplifies nicely for all points xx; this is a collection of those functions.

Simple Rules

ddx[c]=0\frac{d}{dx}[c]=0

ddx[x]=1\frac{d}{dx}[x]=1

ddx[cf(x)]=cddx[f(x)]\frac{d}{dx}[c*f(x)]=c*\frac{d}{dx}[f(x)]

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]\frac{d}{dx}[f(x)+g(x)]=\frac{d}{dx}[f(x)]+\frac{d}{dx}[g(x)]

Power Rule

ddx[xn]=nx(n1)\frac{d}{dx}[x^n]=nx^{(n-1)}

Product Rule

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)

Quotient Rule

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)g(x)2\frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}

Chain Rule

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))]=f'(g(x))*g'(x)

Natural Number (e) Rules

ddx[ex]=ex\frac{d}{dx}[e^x]=e^x

ddx[ecx]=cecx\frac{d}{dx}[e^{cx}]=ce^{cx}

ddx[ln(x)]=1x\frac{d}{dx}[\ln(x)]=\frac{1}{x}

L’Hopital’s Rule

If limxaf(x)\lim_{ x \to a } f(x) and limxag(x)\lim_{ x \to a } g(x) both approach 00 or both approach ±\pm\infty, then:

limxaf(x)g(x)=limxaf(x)g(x)\lim_{ x \to a } \frac{f(x)}{g(x)}=\lim_{ x \to a }\frac{f'(x)}{g'(x)}

Trigonometric Functions

ddx[sin(x)]=cos(x)\frac{d}{dx}[\sin (x)]=\cos(x)

ddx[cos(x)]=sin(x)\frac{d}{dx}[\cos(x)]=-\sin(x)

ddx[tan(x)]=sec2(x)\frac{d}{dx}[\tan(x)]=\sec^2(x)

ddx[sec(x)]=sec(x)tan(x)\frac{d}{dx}[\sec(x)]=\sec (x)*\tan(x)

ddx[csc(x)]=csc(x)cot(x)\frac{d}{dx}[\csc(x)]=-\csc(x)*\cot(x)

ddx[cot(x)]=csc2(x)\frac{d}{dx}[\cot(x)]=-\csc^2(x)

Inverse Trigonometric Functions

ddx[sin1(x)]=11x2\frac{d}{dx}[\sin^{-1}(x)]=\frac{1}{\sqrt{ 1-x^2 }}

ddx[cos1(x)]=11x2\frac{d}{dx}[\cos^{-1}(x)]=-\frac{1}{\sqrt{ 1-x^2 }}

ddx[tan1(x)]=11+x2\frac{d}{dx}[\tan^{-1}(x)]=\frac{1}{1+x^2}

ddx[csc1(x)]=1xx21\frac{d}{dx}[\csc^{-1}(x)]=-\frac{1}{x\sqrt{ x^2-1 }}

ddx[sec1(x)]=1xx21\frac{d}{dx}[\sec^{-1}(x)]=\frac{1}{x \sqrt{ x^2-1 } }

ddx[cot1(x)]=11+x2\frac{d}{dx}[\cot^{-1}(x)]=-\frac{1}{1+x^2}