Math, asked by zainmd8883, 1 year ago

Find the derivatives w.r.t.x:
\rm \frac{\sin x-x\cos x}{x\sin x+\cos x}

Answers

Answered by sprao534
0
Please see the attachment
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Answered by MaheswariS
0

Answer:

\frac{dy}{dx}=\frac{x^2}{(x\:sinx+cosx)^2}

Step-by-step explanation:

Let

y=\frac{sinx-x\:cosx}{x\:sinx+cosx}=\frac{u}{v}(say)

Applying quotient rule of differentiation

\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

\frac{dy}{dx}=\frac{(x\:sinx+cosx)(cosx-(-x\:sinx+cosx.1))-(sinx-x\:cosx)(x\:cosx+sinx.1-sinx)}{(x\:sinx+cosx)^2}

\frac{dy}{dx}=\frac{(x\:sinx+cosx)(cosx+x\:sinx-cosx)-(sinx-x\:cosx)(x\:cosx+sinx-sinx)}{(x\:sinx+cosx)^2}

\frac{dy}{dx}=\frac{(x\:sinx+cosx)(x\:sinx)-(sinx-x\:cosx)(x\:cosx)}{(x\:sinx+cosx)^2}

\frac{dy}{dx}=\frac{(x^2sin^2x+x\:sinx\:cosx)-(x\:sinx\:cosx-x^2\:cos^2x)}{(x\:sinx+cosx)^2}

\frac{dy}{dx}=\frac{x^2sin^2x+x\:sinx\:cosx-x\:sinx\:cosx+x^2\:cos^2x}{(x\:sinx+cosx)^2}

\frac{dy}{dx}=\frac{x^2(sin^2x+cos^2x)}{(x\:sinx+cosx)^2}

Using

\boxed{sin^2A+cos^2A=1}

\frac{dy}{dx}=\frac{x^2(1)}{(x\:sinx+cosx)^2}

\implies\:\boxed{\bf\frac{dy}{dx}=\frac{x^2}{(x\:sinx+cosx)^2}}

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