Rules of Differentiation

Differentiating from first principles is slow. Each rule is that limit worked out once for one shape of expression. The skill is recognizing the shape: a power, a product, a quotient, or a function inside another function.

What does $\frac{dy}{dx}$dy/dx actually mean?

$\frac{dy}{dx}$dy/dx is the rate at which y changes as x changes, at one instant: the gradient of the tangent at a point. Draw a chord through two points on the curve; its gradient is rise over run. Slide one point toward the other and the chord's gradient settles on one value. That value is the derivative, and it is exact, because it is a limit.

Leibniz wrote it $\frac{dy}{dx}$dy/dx and Lagrange wrote f'(x), read "f prime". On its own, $\frac{d}{dx}$d/dx is an instruction: differentiate whatever follows it.

From first principles, write the rise over a step h, divide by h, simplify, and only then let h go to 0. For $y=x^2$y = x²: $\frac{\left({\left(x+h\right)}^2-x^2\right)}{h}=2x+h$((x + h)² − x²)/h = 2x + h, which goes to 2x.

See The derivative as a limit, Derivative notation and Differentiating from first principles.

Where does a derivative fail to exist?

Wherever the curve has no single tangent. At a corner, the gradients on the two sides differ, so no single value fits. At a vertical tangent, the run is 0, so the gradient has no value. At a jump, there is no chord to shrink.

y = |x| has no derivative at 0. The gradient is −1 on the left and +1 on the right.

What is the power rule?

$x^n$xⁿ differentiates to $n{x}^{n-1}$nxⁿ⁻¹. There are two moves, in this order: the power comes down to the front and multiplies, then the power drops by one.

$x^5\to 5x^4$x⁵ → 5x⁴. $\frac{1}{x}=x^{-1}\to -x^{-2}=-\frac{1}{x^2}$1/x = x⁻¹ → −x⁻² = −1/x². $\sqrt{x}=x^{\frac{1}{2}}\to \left(\frac{1}{2}\right)x^{-\frac{1}{2}}=\frac{1}{\left(2\sqrt{x}\right)}$√x = x^(1/2) → (1/2)x^(−1/2) = 1/(2√x).

Rewrite a reciprocal or a root as a power first, then run the same two moves.

A straight line y = mx + c has gradient m everywhere, so its derivative is m. A constant is a flat line, so its derivative is 0.

The sum rule: differentiate each term separately. The constant multiple rule: a numerical factor stays where it is. So $4x^3-7x+2$4x³ − 7x + 2 differentiates to $12x^2-7$12x² − 7. See The power rule, Differentiating linear functions and The constant multiple and sum rules.

When do you use the product rule?

When two functions of x are multiplied together. If y = uv, then $\frac{dy}{dx}=u{v}^{\prime }+$dy/dx = uv' + vu′: differentiate each factor in turn, keep the other as it is, and add.

$y=x^2\sin x\to x^2\cos x+2x\sin x$y = x² sin x → x² cos x + 2x sin x.

For a fraction $\frac{u}{v}$u/v, the quotient rule: $\frac{dy}{dx}=($dy/dx = (vu′ $-u{v}^{\prime })/v^2$− uv')/v². The numerator is a subtraction, so the order matters.

If the denominator is a single power, use the power rule instead: $\frac{3}{x^2}=3x^{-2}\to -6x^{-3}$3/x² = 3x⁻² → −6x⁻³. See The product rule and The quotient rule.

When do you use the chain rule?

Whenever one function sits inside another. Differentiate the outer function, leaving the inside as it is, then multiply by the derivative of the inside.

1. Read the expression as "do this, then do that". ${\left(3x+1\right)}^5$(3x + 1)⁵ means add 1 to 3x, then raise the result to the fifth power.
2. Differentiate the outer layer, keeping the inside as it stands: $5{\left(3x+1\right)}^4$5(3x + 1)⁴.
3. Multiply by the derivative of the inside, which is 3: $15{\left(3x+1\right)}^4$15(3x + 1)⁴.

$\sin \left(x^2\right)\to \cos \left(x^2\right)\times 2x$sin(x²) → cos(x²) × 2x.

The final multiplication is the step most often left out. In Leibniz notation, $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$dy/dx = dy/du × du/dx: the two rates of change multiply.

See The chain rule.

How do you differentiate trigonometric functions?

sin x differentiates to cos x, and cos x differentiates to −sin x. Four differentiations return to sin x.

This holds only in radians; in degrees, every derivative picks up a factor of $\frac{\pi }{180}$π/180. The proof uses the limit of sin x / x from limits and continuity.

tan x = sin x / cos x, so the quotient rule applies. The numerator is ${\cos }^{2}x+{\sin }^{2}x$cos²x + sin²x, which is 1, and $\frac{1}{{\cos }^{2}x}$1/cos²x is ${\sec }^{2}x$sec²x.

The same route gives $\sec x\to \sec x\tan x$sec x → sec x tan x, $\mathrm{cosec}x\to -\mathrm{cosec}x\cot x$cosec x → −cosec x cot x and $\cot x\to -{\mathrm{cosec}}^{2}x$cot x → −cosec²x. See Differentiating trigonometric functions and Differentiating tan x and the reciprocal ratios.

What is e, and why does it matter?

e is the base whose exponential curve has a gradient equal to its height at every point. The gradient of $2^x$2ˣ is less than its height and the gradient of $3^x$3ˣ is greater; near 2.718 they match exactly.

So $e^x$eˣ differentiates to itself. For ln x, write $e^y=x$eʸ = x and differentiate both sides: $e^y\frac{dy}{dx}=1$eʸ dy/dx = 1, so $\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}$dy/dx = 1/eʸ = 1/x.

Any other base is e in disguise: $a^x=e^{x\ln a}$aˣ = e^(x ln a), and the chain rule gives $a^x\ln a$aˣ ln a. Likewise ${\log }_{a}x=\ln x$logₐ x = ln x / ln a differentiates to $\frac{1}{\left(x\ln a\right)}$1/(x ln a).

See The number e, Differentiating exponentials and logarithms and Differentiating $a^x$aˣ and ${\log }_{a}x$logₐ x.

What if the equation is not solved for y?

Differentiate every term as it stands, then solve for $\frac{dy}{dx}$dy/dx. $x^2+y^2=25$x² + y² = 25 is a circle, and no rearranging writes y as a single function of x.

A y-term differentiates as usual and then gains a factor of $\frac{dy}{dx}$dy/dx, by the chain rule. Collect the $\frac{dy}{dx}$dy/dx terms and solve.

$x^2+y^2=25\to 2x+2y\frac{dy}{dx}=0\to \frac{dy}{dx}=-\frac{x}{y}$x² + y² = 25 → 2x + 2y dy/dx = 0 → dy/dx = −x/y.

The radius to (x, y) has gradient $\frac{y}{x}$y/x, so the tangent is perpendicular to the radius. See Implicit differentiation.

How do you differentiate an inverse function?

Take the reciprocal. An inverse function is the reflection of the original in the line y = x, and reflecting swaps the rise with the run. Read f' at the matching point: if f(a) = b, then ${\left(f^{-1}\right)}^{\prime }\left(b\right)=\frac{1}{f^{\prime }\left(a\right)}$(f⁻¹)'(b) = 1/f'(a).

For ${\sin }^{-1}x$sin⁻¹ x, undo the inverse first: sin y = x. Differentiate both sides: $\cos y\frac{dy}{dx}=1$cos y dy/dx = 1, so $\frac{dy}{dx}=\frac{1}{\cos }y$dy/dx = 1/cos y, and Pythagoras replaces cos y with $\sqrt{1-x^2}$√(1 − x²).

The same moves on tan y = x give $\frac{1}{\left(1+x^2\right)}$1/(1 + x²). Both derivatives are algebraic, so an integral of that shape has an inverse-trigonometric answer.

Differentiate again and you get the second derivative, f'′(x) or $\frac{d^{2}y}{d{x}^2}$d²y/dx²: the rate at which the gradient changes. For $y=x^2$y = x², the gradient 2x differentiates to 2.

A positive second derivative means the curve bends upward, like a bowl, which is the test for maxima and minima in the applications of differentiation. See The derivative of an inverse function, Differentiating ${\sin }^{-1}x$sin⁻¹ x and ${\tan }^{-1}x$tan⁻¹ x and Higher derivatives.

What are the hyperbolic functions?

They are built from exponentials: $\cosh x=\frac{\left(e^x+e^{-x}\right)}{2}$cosh x = (eˣ + e⁻ˣ)/2, $\sinh x=\frac{\left(e^x-e^{-x}\right)}{2}$sinh x = (eˣ − e⁻ˣ)/2, and tanh x = sinh x / cosh x.

A chain hanging under its own weight follows a cosh curve, called a catenary.

Square the two definitions and subtract: ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1. Compare ${\cos }^2\theta +{\sin }^2\theta =1$cos²θ + sin²θ = 1, which puts $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ) on a circle; with one sign changed, (cosh t, sinh t) lies on a hyperbola.

sinh x differentiates to cosh x, and cosh x differentiates to sinh x, with no minus sign. tanh x differentiates to $\frac{1}{{\cosh }^{2}x}$1/cosh²x.

The inverses are logarithms. Write x = sinh y, multiply every term by $e^y$eʸ to get a quadratic in $e^y$eʸ, solve it, reject the negative root, and take logarithms: arsinh $x=\ln \left(x+\sqrt{x^2+1}\right)$x = ln(x + √(x² + 1)). arcosh keeps only $x\ge 0$x ≥ 0, because cosh takes each value above 1 twice. See sinh, cosh and tanh, The identity ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1, Differentiating sinh x and cosh x, Inverse hyperbolic functions and Inverse hyperbolics as logarithms.

The mistakes worth naming

• Forgetting the inner derivative. The chain rule ends with a multiplication.
• Differentiating a product factor by factor. The derivative of uv is not u'v'.
• Reversing the quotient rule's numerator. vu′ comes first.
• Using degrees. The derivative of sin x is cos x only in radians.
• Rearranging before differentiating implicitly. No curve needs it, and some cannot be rearranged.

Learn this properly in the app

Every lesson linked above is a screen in Math Challenge with a diagram, a worked example and practice questions. The limit behind these rules is in limits and continuity, using a derivative is the applications of differentiation, and running each rule backward is techniques of integration.