$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x).f'(x) - f(x). g'(x)}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{u}{v}] = \frac{v\frac{\text{d}}{\text{d}x}(u) - u\frac{\text{d}}{\text{d}x}(v)}{[v]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{u}{v}] = \frac{v.u ' - u. v '}{v^{2}}$
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)}
}$
$= \frac{\lim_{\triangle x \rightarrow 0}\frac{g(x)[f(x + \triangle x)-f(x)]}{\triangle x}-\lim_{\triangle x \rightarrow 0}\frac{f(x)[g(x + \triangle x)-g(x)]}{\triangle x}}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + \triangle x)}}$
$= \frac{g(x)\lim_{\triangle x \rightarrow 0}\frac{[f(x + \triangle x)-f(x)]}{\triangle x}-f(x)\lim_{\triangle x \rightarrow 0}\frac{[g(x + \triangle x)-g(x)]}{\triangle x}}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + \triangle x)}}$
$= \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)]-f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{\lim_{\triangle x \rightarrow 0}{g(x).g(x + 0)}}$
$= \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)]-f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{{[g(x)]^{2}}}$
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}]=\frac{g(x).f '(x) - f(x). g'(x)}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{5x -1}{x^{2} +1}] = \frac{(x^{2} +1)\frac{\text{d}}{\text{d}x}[5x -1] - (5x -1)\frac{\text{d}}{\text{d}x}[(x^{2} +1]}{(x^{2} +1)^{2}}$
$= \frac{(x^{2} +1)(5) - (5x -1)(2x) }{(x^{2} +1)^{2}}$
$= \frac{5x^{2} + 5 - (10x^{2} -2x}{(x^{2} +1)^{2}}$
$= \frac{5x^{2} + 5 - 10x^{2} + 2x}{(x^{2} +1)^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{5x -1}{x^{2} +1}] = \frac{-5x^{2} + 2x + 5}{(x^{2} +1)^{2}}$
2) Find the derivative of y= $\frac{x^{2}}{sin(x)}$
Solution : According to the quotient rule for differentiation,
$\frac{\text{d}}{\text{d}x}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{\text{d}}{\text{d}x}[f(x)] - f(x)\frac{\text{d}}{\text{d}x}[g(x)]}{[g(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{x^{2}}{sin(x)}] = \frac{(sin(x))\frac{\text{d}}{\text{d}x}[x^{2}] - x^{2}\frac{\text{d}}{\text{d}x}[sin(x)]}{[sin(x)]^{2}}$
$= \frac{sin(x).(2x) - x^{2}.cos(x). }{[(sin(x)]^{2}}$
$= \frac{2x.sin(x)- x^{2}.cos(x)}{[sin(x)]^{2}}$
$\frac{\text{d}}{\text{d}x}[\frac{x^{2}}{sin(x)}] = \frac{2x.sin(x)- x^{2}.cos(x)}{[sin(x)]^{2}}$