How do you differentiate sin(f(x)) and cos(f(x))?

The derivative of sin(f(x)) is f prime of x times cos(f(x)); the derivative of cos(f(x)) is minus f prime of x times sin(f(x)). For example the derivative of sin(2x) is 2 cos(2x), and the derivative of cos(x squared) is minus 2x sin(x squared).

What is the most common chain rule mistake?

Forgetting the inner factor f prime of x. For example the derivative of e to the power 4x is 4 e to the power 4x, not e to the power 4x; and the derivative of (3x plus 2) to the power 5 is 15(3x plus 2) to the power 4, not 5(3x plus 2) to the power 4.

How do you differentiate e to the power f(x)?

The derivative of e to the power f(x) is f prime of x times e to the power f(x). For example the derivative of e to the power x squared is 2x times e to the power x squared.

How do you differentiate sin(f(x)) and cos(f(x))?

The derivative of sin(f(x)) is f prime of x times cos(f(x)); the derivative of cos(f(x)) is minus f prime of x times sin(f(x)). For example the derivative of sin(2x) is 2 cos(2x).

What is the most common chain rule mistake?

Forgetting the inner factor f prime of x. For example the derivative of e to the power 4x is 4 e to the power 4x, not e to the power 4x.

Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

The chain rule

Q: What is the chain rule?

The chain rule differentiates a composite function, a function of a function. If y = g(u) where u = f(x), then dy/dx equals dy/du times du/dx. In words: differentiate the outer function, then multiply by the derivative of the inside.

Q: How do you differentiate a power of a polynomial?

Use d/dx of (f(x)) to the power n equals n times (f(x)) to the power (n minus 1), all multiplied by f prime of x. For example the derivative of (3x squared plus 1) to the power 5 is 30x times (3x squared plus 1) to the power 4.

Q: How do you differentiate ln(f(x))?

The derivative of ln(f(x)) is f prime of x divided by f(x). For example the derivative of ln(4x squared plus 1) is 8x divided by (4x squared plus 1).

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Theory

The chain rule differentiates a composite function — a function of a function. If \(y=g(u)\) where \(u=f(x)\), then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\): differentiate the outer function, then multiply by the derivative of the inside. It extends every standard derivative to powers, exponentials, logarithms and trig composites — the one thing never to forget is the inner factor \(f'(x)\).

A composite function is built in two stages: first \(x\mapsto u=f(x)\) (the inner function), then \(u\mapsto y=g(u)\) (the outer function). For example \(y=(3x^2+1)^5\) has inner \(u=3x^2+1\) and outer \(y=u^5\); \(y=e^{x^2}\) has inner \(x^2\) and outer \(e^{u}\).

The chain rule says the overall rate of change is the product of the rate of each stage: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). In practice this means: differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inside, \(f'(x)\).

Applied to the standard functions it gives \(\dfrac{d}{dx}(f)^n=n(f)^{n-1}f'\), \(\dfrac{d}{dx}e^{f}=f'e^{f}\), \(\dfrac{d}{dx}\ln f=\dfrac{f'}{f}\), \(\dfrac{d}{dx}\sin f=f'\cos f\) and \(\dfrac{d}{dx}\cos f=-f'\sin f\). Answers should be written in simplest and factorised form.

Key idea. Differentiate the outside, then multiply by the derivative of the inside: \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). Never drop the inner factor \(f'(x)\).

A composite differentiates stage by stage: multiply \(\dfrac{dy}{du}\) by \(\dfrac{du}{dx}\)

A tangent gradient from the chain rule: \(\dfrac{d}{dx}\sin(2x)=2\cos(2x)=2\) at \(x=0\)

The chain rule (Leibniz form) for a composite \(y=g(u)\), \(u=f(x)\):

\[\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\]

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

Power of a function (reduce the index, then multiply by \(f'\)):

\[\frac{d}{dx}\big(f(x)\big)^{n}=n\big(f(x)\big)^{n-1}\,f'(x)\]

\frac{d}{d x} {(f)}^{n} = n {(f)}^{n - 1} f^{'}

Exponential and logarithmic composites:

\[\frac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}\qquad \frac{d}{dx}\ln\big(f(x)\big)=\frac{f'(x)}{f(x)}\]

\frac{d}{d x} e^{f} = f^{'} e^{f}

Trigonometric composites:

\[\frac{d}{dx}\sin\big(f(x)\big)=f'(x)\cos\big(f(x)\big)\qquad \frac{d}{dx}\cos\big(f(x)\big)=-f'(x)\sin\big(f(x)\big)\]

\frac{d}{d x} \sin (f) = f^{'} \cos (f)

Factorise. After applying the rule, take out any common factor, e.g. \(\dfrac{d}{dx}(e^{x}+1)^{3}=3e^{x}(e^{x}+1)^{2}\).

How to apply the chain rule

Spot the composite. Identify the inner function \(u=f(x)\) and the outer function \(y=g(u)\).
Differentiate the outer. Find \(\dfrac{dy}{du}\), leaving the inside unchanged (e.g. \(\dfrac{d}{du}u^{n}=nu^{n-1}\), \(\dfrac{d}{du}e^{u}=e^{u}\)).
Differentiate the inner. Find \(\dfrac{du}{dx}=f'(x)\) — this is the factor most often forgotten.
Multiply and factorise. \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\); write \(u\) back in terms of \(x\) and take out any common factor.

Tangent check. To find a gradient or tangent at \(x=a\), differentiate with the chain rule, then substitute \(x=a\) into \(\dfrac{dy}{dx}\).

Example 1 — power of a polynomial

Differentiate \(y=(3x^2+1)^5\).

Solution

Composite: inner \(u=3x^2+1\), outer \(y=u^{5}\).

Differentiate the inner:

\(\dfrac{du}{dx}\)

\(=\)

\(6x\)

Differentiate the outer:

\(\dfrac{dy}{du}\)	\(=\)	\(5u^{4}\)
	\(=\)	\(5(3x^2+1)^{4}\)

Multiply and factorise, using \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\):

\(\dfrac{dy}{dx}\)	\(=\)	\(5(3x^2+1)^{4}\times 6x\)
	\(=\)	\(30x(3x^2+1)^{4}\)

\(\therefore\ \dfrac{dy}{dx}=30x(3x^2+1)^{4}\)

30 x {(3 x^{2} + 1)}^{4}

Example 2 — exponential \(e^{f(x)}\)

Differentiate \(y=e^{x^2}\).

Solution

Composite: inner \(u=x^{2}\), outer \(y=e^{u}\).

Differentiate the inner:

\(\dfrac{du}{dx}\)

\(=\)

\(2x\)

Differentiate the outer (the exponential is unchanged):

\(\dfrac{dy}{du}\)	\(=\)	\(e^{u}\)
	\(=\)	\(e^{x^2}\)

Multiply, using \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\):

\(\dfrac{dy}{dx}\)	\(=\)	\(e^{x^2}\times 2x\)
	\(=\)	\(2x\,e^{x^2}\)

\(\therefore\ \dfrac{dy}{dx}=2x\,e^{x^2}\)

2 x e^{x^{2}}

Example 3 — logarithm \(\ln(f(x))\)

Differentiate \(y=\ln(4x^2+1)\).

Solution

Composite: inner \(u=4x^{2}+1\), outer \(y=\ln u\).

Differentiate the inner:

\(\dfrac{du}{dx}\)

\(=\)

\(8x\)

Differentiate the outer, using \(\dfrac{d}{du}\ln u=\dfrac{1}{u}\):

\(\dfrac{dy}{du}\)	\(=\)	\(\dfrac{1}{u}\)
	\(=\)	\(\dfrac{1}{4x^2+1}\)

Multiply, using \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\):

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{4x^2+1}\times 8x\)
	\(=\)	\(\dfrac{8x}{4x^2+1}\)

\(\therefore\ \dfrac{dy}{dx}=\dfrac{8x}{4x^2+1}\)

\frac{8 x}{4 x^{2} + 1}

Example 4 — trig composite & a tangent

Find the gradient of \(y=\sin(2x)\) at \(x=0\), and the tangent there.

Solution

Composite: inner \(u=2x\), outer \(y=\sin u\).

Differentiate the inner:

\(\dfrac{du}{dx}\)

\(=\)

\(2\)

Differentiate the outer, using \(\dfrac{d}{du}\sin u=\cos u\):

\(\dfrac{dy}{du}\)	\(=\)	\(\cos u\)
	\(=\)	\(\cos(2x)\)

Multiply for \(\dfrac{dy}{dx}\):

\(\dfrac{dy}{dx}\)	\(=\)	\(\cos(2x)\times 2\)
	\(=\)	\(2\cos(2x)\)

Gradient — substitute \(x=0\):

\(\left.\dfrac{dy}{dx}\right\|_{x=0}\)	\(=\)	\(2\cos 0\)
	\(=\)	\(2\)

Tangent — through \((0,0)\) with gradient \(2\), use \(y-y_{1}=m(x-x_{1})\):

\(y-0\)	\(=\)	\(2(x-0)\)
\(y\)	\(=\)	\(2x\)

\(\therefore\) Gradient \(=2\); tangent: \(y=2x\)

y = 2 x

Common pitfalls

Do not drop the inner factor. \(\dfrac{d}{dx}e^{4x}=4e^{4x}\), not \(e^{4x}\); \(\dfrac{d}{dx}(3x+2)^{5}=15(3x+2)^{4}\), not \(5(3x+2)^{4}\). The derivative of the inside always multiplies.

Reduce the index by one. For a power, use \(n(f)^{n-1}\), not \(n(f)^{n}\): \(\dfrac{d}{dx}(2x-1)^{4}=8(2x-1)^{3}\).

Mind the \(\cos\) sign. \(\dfrac{d}{dx}\cos\big(f(x)\big)=-f'(x)\sin\big(f(x)\big)\); the minus sign is part of the rule.

Frequently asked questions

What is the chain rule?

It differentiates a composite \(y=g(u)\), \(u=f(x)\): \(\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}\). Differentiate the outer function, then multiply by the derivative of the inside.

How do you differentiate a power of a polynomial?

Use \(\dfrac{d}{dx}(f)^n=n(f)^{n-1}f'\). For example \(\dfrac{d}{dx}(3x^2+1)^5=5(3x^2+1)^4\times 6x=30x(3x^2+1)^4\).

How do you differentiate \(e^{f(x)}\)?

\(\dfrac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}\): the exponential is unchanged and you multiply by the derivative of the exponent, e.g. \(\dfrac{d}{dx}e^{x^2}=2x\,e^{x^2}\).

How do you differentiate \(\ln(f(x))\)?

\(\dfrac{d}{dx}\ln(f(x))=\dfrac{f'(x)}{f(x)}\), e.g. \(\dfrac{d}{dx}\ln(4x^2+1)=\dfrac{8x}{4x^2+1}\).

How do you differentiate \(\sin\) and \(\cos\) of a function?

\(\dfrac{d}{dx}\sin(f)=f'\cos(f)\) and \(\dfrac{d}{dx}\cos(f)=-f'\sin(f)\), e.g. \(\dfrac{d}{dx}\sin(2x)=2\cos(2x)\).

What is the most common mistake?

Forgetting the inner factor \(f'(x)\). Always multiply by the derivative of the inside, and then factorise the result.

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