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# Differentiate

Question:

Differentiate

$\left(\frac{\sin x-x \cos x}{x \sin x+\cos x}\right)$

Solution:

To find: Differentiation of $\frac{(\sin x-x \cos x)}{(x \sin x+\cos x)}$

Formula used: $(i)\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sin x}{d x}=\cos x$

(iii) $\frac{d \cos x}{d x}=-\sin x$

(iv) (uv)′ = u′v + uv′ (Leibnitz or product rule)

Let us take $u=(\sin x-x \cos x)$ and $v=(x \sin x+\cos x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d[\sin x-x \cos x]}{d x}$

Applying Product rule for finding the term xcosx in u’

(gh)′ = g′h + gh′

Taking g = xand h = cosx

${[x \cos x]^{\prime}=(1)(\cos x)+x(-\sin x) }$

${[x \cos x]^{\prime}=\cos x-x \sin x }$

Applying the above obtained value for finding u’

$u^{\prime}=\cos x-(\cos x-x \sin x)$

$u^{\prime}=x \sin x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(x \sin x+\cos x)}{d x}$

Applying Product rule for finding the term xsinx in v’

(gh)′ = g′h + gh′

Taking g = xand h = sinx

${[x \sin x]^{\prime}=(1)(\sin x)+x(\cos x) }$

${[x \sin x]^{\prime}=\sin x+x \cos x }$

Applying the above obtained value for finding v’

$v^{\prime}=\sin x+x \cos x-\sin x$

$v^{\prime}=x \cos x$

Putting the above obtained values in the formula:-

$\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{u}^{\prime} \mathrm{v}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ where $\mathrm{v} \neq 0$ (Quotient rule)

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

$=\frac{\left(x^{2} \sin ^{2} x+x \sin x \cos x\right)-\left(x \sin x \cos x-x^{2} \cos ^{2} x\right)}{(x \sin x+\cos x)^{2}}$

$=\frac{\left.x^{2} \sin ^{2} x+x \sin x \cos x-x \sin x \cos x+x^{2} \cos ^{2} x\right)}{(x \sin x+\cos x)^{2}}$

$=\frac{x^{2}\left(\sin ^{2} x+\cos ^{2} x\right)}{(x \sin x+\cos x)^{2}}$

$=\frac{x^{2}}{(x \sin x+\cos x)^{2}}$

Ans) $=\frac{x^{2}}{(x \sin x+\cos x)^{2}}$