Differentiation

These are the standard rules for computing derivatives of functions f,g:R→Rf, g : \mathbb{R} \to \mathbb{R}. They are the basic toolkit used whenever partial derivatives are computed component-by-component.

Core Rules

Let ff and gg be differentiable functions and c∈Rc \in \mathbb{R} a constant. The fundamental differentiation rules are:

RuleFormula
Constantddx[c]=0\frac{d}{dx}[c] = 0
Powerddx[xn]=nxn−1\frac{d}{dx}[x^n] = n x^{n-1}
Constant multipleddx[c⋅f]=c⋅f′\frac{d}{dx}[c \cdot f] = c \cdot f'
Sumddx[f+g]=f′+g′\frac{d}{dx}[f + g] = f' + g'
Productddx[f⋅g]=f′g+fg′\frac{d}{dx}[f \cdot g] = f' g + f g'
Quotientddx ⁣[fg]=f′g−fg′g2\frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f' g - f g'}{g^2}
Chainddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

The last rule is the chain rule — it is the most important rule for computing partial derivatives of composite functions.

Common Derivatives

f(x)f(x)f′(x)f'(x)
cc00
xnx^nnxn−1n x^{n-1}
exe^xexe^x
eaxe^{ax}aeaxa e^{ax}
ln⁡x\ln x1x\dfrac{1}{x}
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
tan⁡x\tan x1cos⁡2x\dfrac{1}{\cos^2 x}

Trigonometric Derivatives

The trigonometric derivatives follow from the rules above combined with the limit definitions of sin⁡\sin and cos⁡\cos. The full set:

The derivatives of the trigonometric functions are:

f(x)f(x)f′(x)f'(x)
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
tan⁡x\tan x1cos⁡2x=sec⁡2x\dfrac{1}{\cos^2 x} = \sec^2 x
cot⁡x\cot x−1sin⁡2x=−csc⁡2x-\dfrac{1}{\sin^2 x} = -\csc^2 x
sec⁡x\sec xsec⁡xtan⁡x\sec x \tan x
csc⁡x\csc x−csc⁡xcot⁡x-\csc x \cot x

For the inverse trigonometric functions:

ddxarcsin⁡x=11−x2ddxarccos⁡x=−11−x2ddxarctan⁡x=11+x2\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}} \qquad \frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}} \qquad \frac{d}{dx} \arctan x = \frac{1}{1 + x^2}

Chain Rule in Detail

The chain rule handles composite functions. If h(x)=f(g(x))h(x) = f(g(x)), then:

h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)

Let h(x)=e−xsin⁡xh(x) = e^{-x} \sin x. Using the product rule:

h′(x)=(e−x)′sin⁡x+e−x(sin⁡x)′=−e−xsin⁡x+e−xcos⁡x=e−x(cos⁡x−sin⁡x)h'(x) = (e^{-x})' \sin x + e^{-x} (\sin x)' = -e^{-x} \sin x + e^{-x} \cos x = e^{-x}(\cos x - \sin x)

Let h(x)=sin⁡(x2)h(x) = \sin(x^2). Using the chain rule with f(u)=sin⁡uf(u) = \sin u and g(x)=x2g(x) = x^2:

h′(x)=cos⁡(x2)⋅2x=2xcos⁡(x2)h'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2)

Translation Invariance

For any fixed constant a∈Ra \in \mathbb{R}, the shifted variable (x−a)(x - a) behaves under differentiation exactly like xx alone — a property called translation invariance:

ddx(x−a)=1\frac{d}{dx}(x - a) = 1

This follows from the constant rule and sum rule together: ddx[x−a]=ddx[x]−ddx[a]=1−0=1\frac{d}{dx}[x - a] = \frac{d}{dx}[x] - \frac{d}{dx}[a] = 1 - 0 = 1. As a consequence, all the standard rules apply to expressions in (x−a)(x - a) exactly as they apply to expressions in xx:

ddx[(x−a)n]=n(x−a)n−1ddx[f(x−a)]=f′(x−a)\frac{d}{dx}[(x - a)^n] = n(x - a)^{n-1} \qquad \frac{d}{dx}[f(x - a)] = f'(x - a)

The same holds in higher dimensions: for fixed a∈Rn\mathbf{a} \in \mathbb{R}^n, the Jacobian of the shift is the identity matrix,

∂∂x(x−a)=I\frac{\partial}{\partial \mathbf{x}}(\mathbf{x} - \mathbf{a}) = I

so derivatives of expressions in (x−a)(\mathbf{x} - \mathbf{a}) behave exactly like derivatives of expressions in x\mathbf{x} alone.

This is what makes Taylor expansions around any base point — in powers of (x−a)(x - a) or (x−a)(\mathbf{x} - \mathbf{a}) — no harder to manipulate than expansions around the origin.

Partial Derivatives via These Rules

When computing partial derivatives, all the rules above apply — simply treat every variable other than the one being differentiated as a constant. For example, with f(x,y,z)=e−zsin⁡x+y2f(x, y, z) = e^{-z} \sin x + y^2:

  • ∂f∂x\frac{\partial f}{\partial x}: treat y,zy, z as constants → e−zcos⁡xe^{-z} \cos x
  • ∂f∂y\frac{\partial f}{\partial y}: treat x,zx, z as constants → 2y2y
  • ∂f∂z\frac{\partial f}{\partial z}: treat x,yx, y as constants, apply chain rule on e−ze^{-z} → −e−zsin⁡x-e^{-z} \sin x