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 such that

Moreover, the derivative of 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,

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

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

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

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

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

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