Differentiation rules

The first principle of calculus is tedious and it gives you plenty of opportunities to make mistakes. Fortunately, there are some shortcuts.

Constant rule

This is really just a special case of the product rule, but you will use it so often that it’s easier to think of it separately. If k\displaystyle k is a constant, then

ddx(kf(x))=kddxf(x)\displaystyle \frac{d}{dx} \left ( k \cdot f{\left ( x \right )} \right ) = k \cdot \frac{d}{dx} f{\left ( x \right )}.

Sum & difference rules

The derivative of the sum is equal to the sum of the derivatives. The same goes for subtraction. This means that

(f±g)=f±g\displaystyle \left ( f \pm g \right ) ' = f{'} \pm g{'}.

or, using Leibniz’s notation,

ddx(u±v)=dudx±dvdx\displaystyle \frac{d}{dx} \left ( u \pm v \right ) = \frac{d u}{dx} \pm \frac{d v}{dx}.

Power rule

Polynomials are very easy to differentiate thanks to the power rule.

ddx(xn)=nxn1\displaystyle \frac{d}{dx} \left ( x^{n} \right ) = n x^{n - 1}.

For example, the derivative of x100\displaystyle x^{100} with respect to x\displaystyle x is 100x99\displaystyle 100 x^{99}.

Product rule

The derivative of the product is not the product of the derivatives. Rather,

(fg)=fg+fg\displaystyle \left ( f{g} \right ) ' = f{'} g + f{g{'}},

or, using Leibniz’s notation,

ddx(uv)=dudxv+dvdxu\displaystyle \frac{d}{dx} \left ( u v \right ) = \frac{d u}{dx} \cdot v + \frac{d v}{dx} \cdot u.

Quotient rule

We can use the product and power rules together to tackle rationals, but using the quotient rule is much easier:

(fg)=fgfgg2\displaystyle \left ( \frac{f}{g} \right ) ' = \frac{f{'} g - f{g{'}}}{g^{2}},

or, using Leibniz’s notation,

ddx(uv)=v2(dudxvdvdxu)\displaystyle \frac{d}{dx} \left ( \frac{u}{v} \right ) = v^{- 2} \left ( \frac{d u}{dx} \cdot v - \frac{d v}{dx} \cdot u \right ).

Chain rule

To differentiate composite functions, we need to use the chain rule:

(fg)=(fg)g\displaystyle \left ( f \circ g \right ) ' = \left ( f{'} \circ g \right ) \cdot g{'},

or, using Leibniz’s notation,

dydx=dydududx\displaystyle \frac{dy}{dx} = \frac{dy}{d u} \cdot \frac{d u}{dx}.