Question

Differentiate the function w.r.t.x:
$\rm \frac{x^{2}+4}{x\log x+5}$

Answer 1

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Answer 2

We know that such form of expression can solve by U/V method of differentiation

$\frac{d}{dx} \bigg(\frac{U}{V}\bigg ) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} } \\ \\ here \: U = x^{2}+4 \\ \\ V = x\log x+5$
$\frac{d}{dx} \bigg(\frac{x^{2}+4}{x\log x+5} \bigg) = \frac{( x\log x+5) \frac{d(x^{2}+4)}{dx} -(x^{2}+4)\frac{d( x\log x+5)}{dx} }{( { x\log x+5})^{2} }$

$\frac{d}{dx} \bigg(\frac{x^{2}+4}{x\log x+5} \bigg) = \frac{(x\log x+5)(2x) -(x^{2}+4)(x(\frac{1}{x})+log\:x(1) }{( {x\log x+5})^{2} } \\ \\ \frac{d}{dx} \bigg(\frac{x^{2}+4}{x\log x+5} \bigg)= \frac{(x\log x+5)(2x) -(x^{2}+4)(1+log\:x)}{( {x\log x+5})^{2} } \\ \\ \\ \frac{d}{dx} \bigg(\frac{x^{2}+4}{x\log x+5} \bigg)= \frac{(2x^{2}\log x+10x -x^{2}-x^{2}\log x-4-4log\:x)}{( {x\log x+5})^{2} } \\ \\\frac{d}{dx} \bigg(\frac{x^{2}+4}{x\log x+5} \bigg)=\frac{(x^{2}\log x+10x -x^{2}-4(1+log\:x)}{( { x\log x+5})^{2} }$