Differentiation

A progressive guide to derivatives, from the limit definition and core differentiation rules to implicit differentiation, inverse functions, and higher derivatives.

What a measures

Differentiation is the process of finding a . If a function is written as y=f(x)y=f(x), common notations for its are

f′(x)=dydx=ddx[f(x)].f'(x)=\frac{dy}{dx}=\frac{d}{dx}[f(x)].

The has two closely related interpretations:

  • Geometrically, it is the slope of the tangent line to a curve at a point.

  • In an application, it is the instantaneous rate of change of one quantity with respect to another.

For example, if position is a function of time, its is instantaneous velocity. The is therefore a local measurement: it describes what is happening near a particular input rather than over a broad interval.

Takeaway: Differentiation converts a function into a new function that records local slope or instantaneous change.

The from a limit

A secant line passes through two points on the graph of y=f(x)y=f(x). If the inputs differ by hh, its slope is the

f(x+h)−f(x)h,h≠0.\frac{f(x+h)-f(x)}{h}, \qquad h\ne 0.

As the second point moves toward the first, h→0h\to 0, the secant slope approaches the slope of the tangent line. This gives the limit definition of the :

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.

An equivalent form at the input aa is

f′(a)=lim⁡x→af(x)−f(a)x−a.f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.

The exists at x=ax=a only when this limit exists as a finite number. At an endpoint, a one-sided limit may be used.

Example: Finding the of f(x)=x2f(x)=x^2

f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0x2+2xh+h2−x2h=lim⁡h→0(2x+h)=2x.\begin{aligned} f'(x) &=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &=\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h}\\ &=\lim_{h\to 0}(2x+h)\\ &=2x. \end{aligned}

Thus, the slope at x=ax=a is 2a2a, and the tangent line at (a,a2)(a,a^2) is

y−a2=2a(x−a).y-a^2=2a(x-a).

Takeaway: The is obtained by simplifying the average-rate expression and then taking its limit as the input change approaches zero.

Differentiability and continuity

Differentiability and continuity are related but not equivalent. If a function is differentiable at aa, then it is continuous at aa. However, continuity alone does not guarantee differentiability.

A function can fail to be differentiable at a:

  • corner,

  • cusp,

  • vertical tangent, or

  • discontinuity.

For example, f(x)=∣x∣f(x)=|x| is continuous at 00, but it has no there. The slope from the left is −1-1, while the slope from the right is 11; because the one-sided slopes do not agree, the does not exist.

Takeaway: Differentiability implies continuity, but a continuous function may still have a sharp feature where no single tangent slope exists.

Core differentiation rules

Basic rules make it possible to differentiate complicated expressions efficiently.

  • :

    ddx[xn]=nxn−1.\frac{d}{dx}[x^n]=nx^{n-1}.
  • Constant rule:

    ddx[c]=0.\frac{d}{dx}[c]=0.
  • Constant-multiple rule:

    ddx[cf(x)]=cf′(x).\frac{d}{dx}[cf(x)]=cf'(x).
  • Sum and difference rules:

    ddx[f(x)±g(x)]=f′(x)±g′(x).\frac{d}{dx}[f(x)\pm g(x)]=f'(x)\pm g'(x).
  • :

    ddx[uv]=u′v+uv′.\frac{d}{dx}[uv]=u'v+uv'.
  • :

    ddx[uv]=vu′−uv′v2.\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{vu'-uv'}{v^2}.

For example,

ddx[x2sin⁡x]=2xsin⁡x+x2cos⁡x.\frac{d}{dx}[x^2\sin x]=2x\sin x+x^2\cos x.

For an expression that is a composition, use the :

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

The generalized is a common chain-rule application:

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

Takeaway: First identify the structure of the expression, then select the rule that matches that structure.

Algebraic functions

Rewrite radicals and reciprocals as powers when this makes the rules easier to apply:

x=x1/2,1x4=x−4.\sqrt{x}=x^{1/2}, \qquad \frac{1}{x^4}=x^{-4}.

For example,

ddx(3x5−4x+2x3)=15x4−2x−1/2−6x−4.\frac{d}{dx}\left(3x^5-4\sqrt{x}+\frac{2}{x^3}\right) =15x^4-2x^{-1/2}-6x^{-4}.

Algebraic functions may also require several rules at once. Consider

f(x)=(x2+1)3x=(x2+1)3x−1/2.f(x)=\frac{(x^2+1)^3}{\sqrt{x}}=(x^2+1)^3x^{-1/2}.

Using the and the gives

f′(x)=6x(x2+1)2x−1/2−12(x2+1)3x−3/2.f'(x)=6x(x^2+1)^2x^{-1/2}-\frac12(x^2+1)^3x^{-3/2}.

A factored is often useful because it can make zeros, factors, and sign changes easier to see. Always keep domain restrictions in mind when radicals occur in denominators or when negative powers are used.

Takeaway: Converting algebraic expressions to powers often exposes the product, quotient, and composition structure needed for differentiation.

Trigonometric, exponential, and logarithmic derivatives

When differentiating trigonometric, exponential, or logarithmic functions, combine the relevant basic formula with the when the input is not simply xx. Trigonometric formulas assume angles are measured in radians.

Trigonometric functions

ddx[sin⁡x]=cos⁡x,ddx[cos⁡x]=−sin⁡x,ddx[tan⁡x]=sec⁡2x,ddx[cot⁡x]=−csc⁡2x,ddx[sec⁡x]=sec⁡xtan⁡x,ddx[csc⁡x]=−csc⁡xcot⁡x.\begin{aligned} \frac{d}{dx}[\sin x]&=\cos x, & \frac{d}{dx}[\cos x]&=-\sin x, & \frac{d}{dx}[\tan x]&=\sec^2x,\\ \frac{d}{dx}[\cot x]&=-\csc^2x, & \frac{d}{dx}[\sec x]&=\sec x\tan x, & \frac{d}{dx}[\csc x]&=-\csc x\cot x. \end{aligned}

For example,

ddx[cos⁡(4x3−2x)]=−sin⁡(4x3−2x)(12x2−2).\frac{d}{dx}[\cos(4x^3-2x)]=-\sin(4x^3-2x)(12x^2-2).

Exponential functions

ddx[eu(x)]=eu(x)u′(x),\frac{d}{dx}[e^{u(x)}]=e^{u(x)}u'(x),

and for a>0a>0, a≠1a\ne 1,

ddx[au(x)]=au(x)ln⁡(a)u′(x).\frac{d}{dx}[a^{u(x)}]=a^{u(x)}\ln(a)u'(x).

Thus,

ddx[52x2+1]=52x2+1ln⁡(5)(4x).\frac{d}{dx}[5^{2x^2+1}]=5^{2x^2+1}\ln(5)(4x).

Logarithmic functions

ddx[ln⁡∣u(x)∣]=u′(x)u(x),\frac{d}{dx}[\ln|u(x)|]=\frac{u'(x)}{u(x)},

where u(x)≠0u(x)\ne 0. In particular,

ddx[ln⁡(3x2+1)]=6x3x2+1.\frac{d}{dx}[\ln(3x^2+1)]=\frac{6x}{3x^2+1}.

For a logarithm with base aa,

ddx[log⁡ax]=1xln⁡a.\frac{d}{dx}[\log_a x]=\frac{1}{x\ln a}.

Takeaway: Memorize the basic families, then multiply by the of the inner expression whenever the input is composite.

Inverse functions

The of an inverse function can be found from the of the original function. If ff is one-to-one and differentiable, then

(f−1)′(x)=1f′(f−1(x)),(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))},

provided the denominator is nonzero.

Important inverse trigonometric formulas include

ddx[arcsin⁡x]=11−x2,ddx[arccos⁡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}},

and

ddx[arctan⁡x]=11+x2.\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}.

With a composite argument,

ddx[arctan⁡(u)]=u′1+u2.\frac{d}{dx}[\arctan(u)]=\frac{u'}{1+u^2}.

For example,

ddx[arctan⁡(3x2)]=6x1+9x4.\frac{d}{dx}[\arctan(3x^2)]=\frac{6x}{1+9x^4}.

The formula for arcsin⁡x\arcsin x can be derived by setting y=arcsin⁡xy=\arcsin x, so that sin⁡y=x\sin y=x. Differentiating implicitly gives cos⁡y dy/dx=1\cos y\,dy/dx=1, and the principal range of arcsin⁡\arcsin supplies cos⁡y=1−x2\cos y=\sqrt{1-x^2}.

Takeaway: Inverse-function derivatives are reciprocals of the corresponding original-function derivatives, evaluated at the matching inverse input.

Finding slopes implicitly

Use when an equation relates xx and yy but does not conveniently give yy as an explicit function of xx. Differentiate both sides with respect to xx, treat yy as a function of xx, apply the to terms involving yy, and solve for dy/dxdy/dx.

For example, given the circle

x2+y2=25,x^2+y^2=25,

differentiation gives

2x+2ydydx=0.2x+2y\frac{dy}{dx}=0.

Therefore,

dydx=−xy.\frac{dy}{dx}=-\frac{x}{y}.

At (3,4)(3,4), the slope is −3/4-3/4, so the tangent line is

y−4=−34(x−3).y-4=-\frac34(x-3).

For a more complicated relation,

x2+xy+y2=7,x^2+xy+y^2=7,

differentiation gives

2x+(xdydx+y)+2ydydx=0.2x+\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0.

Collecting the terms yields

dydx=−2x+yx+2y.\frac{dy}{dx}=-\frac{2x+y}{x+2y}.

Takeaway: Every occurrence of yy must be differentiated as a function of xx; this is why factors of dy/dxdy/dx appear.

and a differentiation strategy

A second is the of the first :

f′′(x)=ddx[f′(x)]=d2ydx2.f''(x)=\frac{d}{dx}[f'(x)]=\frac{d^2y}{dx^2}.

More generally, repeated differentiation produces . For

f(x)=x4−3x2+2x,f(x)=x^4-3x^2+2x,

we obtain

f′(x)=4x3−6x+2,f′′(x)=12x2−6.f'(x)=4x^3-6x+2, \qquad f''(x)=12x^2-6.

The first describes instantaneous rate of change. The second describes how that rate changes. In motion problems, if position is differentiated once to obtain velocity, differentiating again gives acceleration.

For a complicated function, use this workflow:

  1. Identify the outermost operation: sum, product, quotient, or composition.

  2. Rewrite radicals and reciprocals as powers when helpful.

  3. Apply the appropriate structural rule.

  4. Differentiate the basic pieces.

  5. Simplify only when simplification improves clarity.

  6. Check restrictions from logarithms, denominators, radicals, and inverse trigonometric functions.

For example, with

f(x)=ex2ln⁡(sin⁡x),f(x)=e^{x^2}\ln(\sin x),

use the and the :

f′(x)=2xex2ln⁡(sin⁡x)+ex2cot⁡x.f'(x)=2xe^{x^2}\ln(\sin x)+e^{x^2}\cot x.

The real-valued function requires sin⁡x>0\sin x>0, and the is defined only where the entire expression is defined.

Takeaway: A reliable differentiation strategy is structural: identify the outer operation, apply the matching rule, differentiate inward, and verify the domain.