Quotient Rule for Differentiation

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Quotient rule for differentiation
If f(x)=u and g(x)= v are two differentiable functions and g(x) $\ne$ 0 then show that $\frac{f(x)}{g(x)}$ is also differentiable for all values of x.

Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x).f'(x) - f(x). g'(x)}{[g(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{u}{v}] = \frac{v\frac{\text{d}}{\text{d}x}(u) - u\frac{\text{d}}{\text{d}x}(v)}{[v]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{u}{v}] = \frac{v.u ' - u. v '}{v^{2}}$


Proof: Since f(x) and g(x) are differentiable functions.

By definition of derivative,
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \lim_{\triangle x \rightarrow 0}\frac{\frac{f(x + \triangle x}{g(x + \triangle x})-\frac{f(x)}{g(x)}}{{\triangle x}}$

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \lim_{\triangle x \rightarrow 0}\frac{g(x).f(x + \triangle x) - f(x).g(x + \triangle x)}{(\triangle x).g(x +\triangle x).g(x)}$

Add and subtract $f(x).g(x)$

$ {\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \lim_{\triangle x \rightarrow 0}\frac{g(x).f(x + \triangle x) \color{red}{-f(x).g(x) + f(x).g(x)}- f(x).g(x + \triangle x)}{(\triangle x).g(x +\triangle x).g(x)} }$

$= \frac{\lim_{\triangle x \rightarrow 0}\frac{g(x)[f(x + \triangle x)-f(x)]}{\triangle x}-\lim_{\triangle x \rightarrow 0}\frac{f(x)[g(x + \triangle x)-g(x)]}{\triangle x}}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + \triangle x)}}$

$= \frac{g(x)\lim_{\triangle x \rightarrow 0}\frac{[f(x + \triangle x)-f(x)]}{\triangle x}-f(x)\lim_{\triangle x \rightarrow 0}\frac{[g(x + \triangle x)-g(x)]}{\triangle x}}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + \triangle x)}}$

$= \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)]-f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + 0)}}$

$= \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)]-f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{{[g(x)]^{2}}}$

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}]=\frac{g(x).f '(x) - f(x). g'(x)}{[g(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$

Note: $\lim_{\triangle x \rightarrow 0}g(x + \triangle x) = g(x)$ because g(x) is given to be differentiable and continuous function.

Examples on Quotient rule for Differentiation

1) Find the derivative of y= $\frac{5x -1}{x^{2} + 1}$

Solution : According to the quotient rule for differentiation,

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{5x -1}{x^{2} +1}] = \frac{(x^{2} +1)\frac{\text{d}}{\text{d}x}[5x -1] - (5x -1)\frac{\text{d}}{\text{d}x}[(x^{2} +1]}{(x^{2} +1)^{2}}$

$= \frac{(x^{2} +1)(5) - (5x -1)(2x) }{(x^{2} +1)^{2}}$

$= \frac{5x^{2} + 5 - (10x^{2} -2x}{(x^{2} +1)^{2}}$

$= \frac{5x^{2} + 5 - 10x^{2} + 2x}{(x^{2} +1)^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{5x -1}{x^{2} +1}] = \frac{-5x^{2} + 2x + 5}{(x^{2} +1)^{2}}$

2) Find the derivative of y= $\frac{x^{2}}{sin(x)}$

Solution : According to the quotient rule for differentiation,

$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{x^{2}}{sin(x)}] = \frac{(sin(x))\frac{\text{d}}{\text{d}x}[x^{2}] - x^{2}\frac{\text{d}}{\text{d}x}[sin(x)]}{[sin(x)]^{2}}$

$= \frac{sin(x).(2x) - x^{2}.cos(x). }{[(sin(x)]^{2}}$

$= \frac{2x.sin(x)- x^{2}.cos(x)}{[sin(x)]^{2}}$

$\frac{\text{d}}{\text{d}x}[\frac{x^{2}}{sin(x)}] = \frac{2x.sin(x)- x^{2}.cos(x)}{[sin(x)]^{2}}$



12th grade math

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