Quotient Rule (Leaving Cert Mathematics): Revision Notes

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Quotient Rule

The quotient rule is a differentiation technique used to find the derivative of a quotient of two functions. It's particularly useful when dealing with functions that are divided by one another, and it complements the product rule and chain rule in calculus.

If a function can be written as :

y=uvy=\frac{u}{v}

then the first derivative can be written as :

dydx=vdudx−udvdxv2\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Example

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Differentiate y=x2+1x2−1y=\frac{x^2+1}{x^2-1}

First, identify the denominator and numerator :

u=x2+1u=x^2+1v=x2−1v=x^2-1

Differentiate each function :

u′=2xu'=2xv′=2xv'=2x

Apply quotient rule :

dydx=(x2−1)⋅2x−(x2+1)⋅2x(x2−1)2\frac{dy}{dx} = \frac{(x^2 - 1) \cdot 2x - (x^2 + 1) \cdot 2x}{(x^2 - 1)^2}

Simplify :

dydx=2x(x2−1−x2−1)(x2−1)2=2x(−2)(x2−1)2=−4x(x2−1)2\frac{dy}{dx} = \frac{2x(x^2 - 1 - x^2 - 1)}{(x^2 - 1)^2} = \frac{2x(-2)}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}

Example

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Differentiate y=exln⁡(x)y= \frac{e^x}{\ln(x)}

Identify the denominator and numerator :

u=exu=e^xv=ln⁡(x)v=\ln{(x)}

Differentiate each function :

u′=exu'=e^xv′=1xv'=\frac{1}{x}

Apply quotient rule :

dydx=ln⁡(x)⋅ex−ex⋅1x[ln⁡(x)]2\frac{dy}{dx} = \frac{\ln(x) \cdot e^x - e^x \cdot \frac{1}{x}}{[\ln(x)]^2}

Simplify :

dydx=ex(ln⁡(x)−1x)[ln⁡(x)]2\frac{dy}{dx} = \frac{e^x \left(\ln(x) - \frac{1}{x}\right)}{[\ln(x)]^2}

Example

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Differentiate y=x2e3xy= \frac{x^2}{e^{3x}}

Identify the denominator and numerator :

u=x2u=x^2v=e3xv=e^{3x}

Differentiate each function :

u′=2xu'=2xv′=3e3xv'=3e^{3x}

Apply quotient rule :

dydx=e3x⋅2x−x2⋅3e3x(e3x)2\frac{dy}{dx} = \frac{e^{3x} \cdot 2x - x^2 \cdot 3e^{3x}}{(e^{3x})^2}

Simplify :

dydx=e3x(2x−3x2)e6x=2x−3x2e3x\frac{dy}{dx} = \frac{e^{3x}(2x - 3x^2)}{e^{6x}} = \frac{2x - 3x^2}{e^{3x}}

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