$\frac{x \tan x}{(\sec x+\tan x)}$



To find: Differentiation of $\frac{x \tan x}{(\sec x+\tan x)}$

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) $\frac{d \sec x}{d x}=\sec x \tan x$

(iii) $\frac{d \tan x}{d x}=\sec ^{2} x$

(iii) $\frac{d x^{n}}{d x}=n x^{n-1}$

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

Let us take u = (x tanx) and v = (secx + tanx)

$u^{\prime}=\frac{d u}{d x}=\frac{d[x \tan x]}{d x}$

Applying Product rule for finding u’

$(\mathrm{gh})^{\prime}=\mathrm{g}^{\prime} \mathrm{h}+\mathrm{gh}^{\prime}$

Taking g = xand h = tanx

$[x \tan x]^{\prime}=(1)(\tan x)+x\left(\sec ^{2} x\right)$

$=\tan x+x \sec ^{2} x$

$u^{\prime}=\tan x+x \sec ^{2} x$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sec x+\tan x)}{d x}=\sec x \tan x+\sec ^{2} x$

$v^{\prime}=\sec x(\tan x+\sec x)$

Putting the above obtained values in the formula:-

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

$\left[\frac{x \tan x}{(\sec x+\tan x)}\right]^{\prime}=\frac{\left(\tan x+x \sec ^{2} x\right)(\sec x+\tan x)-[x \tan x][\sec x(\tan x+\sec x)]}{(\sec x+\tan x)^{2}}$

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

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

$=\frac{\tan x+x \sec x(\sec x-\tan x)}{(\sec x+\tan x)}$

Ans $)=\frac{\tan x+x \sec x(\sec x-\tan x)}{(\sec x+\tan x)}$


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