Differentiation Rules

Differentiate using the chain rule, product rule, and quotient rule at A-Level.

Beyond basic differentiation, A-Level requires the chain rule (for composite functions), product rule, and quotient rule.

The Rules

Chain Rule

dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

y=(3x+1)5y = (3x+1)^5: let u=3x+1u = 3x+1. dydu=5u4\frac{dy}{du} = 5u^4, dudx=3\frac{du}{dx} = 3. dydx=15(3x+1)4\frac{dy}{dx} = 15(3x+1)^4.

Product Rule

ddx[uv]=udvdx+vdudx\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}

y=x2sin⁡xy = x^2\sin x: dydx=x2cos⁡x+2xsin⁡x\frac{dy}{dx} = x^2\cos x + 2x\sin x.

Quotient Rule

ddx[uv]=vdudx−udvdxv2\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

y=x2cos⁡xy = \frac{x^2}{\cos x}: dydx=2xcos⁡x+x2sin⁡xcos⁡2x\frac{dy}{dx} = \frac{2x\cos x + x^2\sin x}{\cos^2 x}.

Standard Derivatives

f(x)f(x) f′(x)f'(x)
sin⁡x\sin x cos⁡x\cos x
cos⁡x\cos x −sin⁡x-\sin x
tan⁡x\tan x sec⁡2x\sec^2 x
exe^x exe^x
ln⁡x\ln x 1x\frac{1}{x}
axa^x axln⁡aa^x \ln a

Worked Example: Chain Rule

Problem

y=e3x2y = e^{3x^2} → dydx=6xe3x2\frac{dy}{dx} = 6xe^{3x^2}.

Solution

Worked Example: Product Rule

Problem

y=x3exy = x^3 e^x → dydx=3x2ex+x3ex=ex(3x2+x3)\frac{dy}{dx} = 3x^2 e^x + x^3 e^x = e^x(3x^2 + x^3).

Solution

Worked Example: Quotient Rule

Problem

y=ln⁡xxy = \frac{\ln x}{x} → dydx=1x⋅x−ln⁡x⋅1x2=1−ln⁡xx2\frac{dy}{dx} = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}.

Solution

Practice Problems

    1. Differentiate y=sin⁡(5x)y = \sin(5x).
    1. Differentiate y=x2ln⁡xy = x^2 \ln x.
    1. Differentiate y=exx+1y = \frac{e^x}{x+1}.

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Key Takeaways

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    Chain rule for composite functions.

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    Product rule for products.

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    Quotient rule for fractions.

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    Memorise standard derivatives.

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