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The Quotient Rule

The Quotient Rule

The quotient rule allows us to directly differentiate functions of the form $$f(x) = \frac{u(x)}{v(x)}$$ We have $$ \frac{d}{dx}\left( \frac{u(x)}{v(x)}\right) = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2} $$

The formula arises as a natural consequence of rewriting $\frac{u}{v}$ as $$ \frac{u}{v} = u\times v^{-1} $$ and then applying the chain and product rules

The Quotient Rule 1

Determine $\frac{dy}{dx}$ for the curve $y = \frac{x^2 + 1}{x+2}$
solution - press button to display

Let $u = x^2 + 1,\;v = x+2$, then $\frac{du}{dx} = 2x,\;\frac{dv}{dx} = 1$

Applying the quotient rule gives $$ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(2x)(x+2) - (1)(x^2+1) }{(x+2)^2} $$

This simplifies to $$ \frac{dy}{dx} = \frac{x^2+4x - 1}{(x+2)^2} $$

The Quotient Rule 2

The quotient rule is of particular importance when differentiating trigonometric functions:

Determine $\frac{dy}{dx}$ for the curve $y =\tan(x)$

solution - press button to display

$$ \tan(x) \equiv \frac{\sin(x)}{\cos(x)} $$

Let $u = \sin(x),\;v = \cos(x)$. Then $\frac{du}{dx} = \cos(x),\;\frac{dv}{dx} = -\sin(x)$

Application of the quotient rule yields $$ \frac{dy}{dx} = \frac{(\cos(x))(\cos(x)) -(\sin(x))(-\sin(x))}{\cos^2(x)} $$

This simplifies to $$ \frac{dy}{dx} = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} = \sec^2(x) $$