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  • Differentiate each of the following w.r.t. x tan inverse sec x + tan x

Differentiate each of the following w.r.t. x
\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{4}<x<\frac{\pi}{2}

Answers (1)

We have given \tan ^{-1}(\sec x+\tan x),-\frac{\pi}{4}<x<\frac{\pi}{2}

Let us Assume (sec x +tan x) =t

So, y = tan^{-1} t

Now, differentiate w.r.t t

\\\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{1+\mathrm{t}^{2}}$ \\since, $\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}$ \\And, $t=(\sec x+\tan x)$ \\Differentiate t w.r.t $x$ \\$\frac{d t}{d x}=\sec x \cdot \tan x+\sec ^{2} x$ \\Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+t^{2}} \times\left(\sec x \tan x+\sec ^{2} x\right)\\ &\text { When, substitute the value of t we get }\\ &\frac{d y}{d x}=\frac{1}{1+(\sec x+\tan x)^{2}} \times\left(\sec x \tan x+\sec ^{2} x\right)\\ &\frac{d y}{d x}=\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \times \sec x \cdot(\sec x+\tan x)\\ &\frac{d y}{d x}=\frac{1}{2 \sec x(\tan x+\sec x)} \times \sec x \cdot(\sec x+\tan x)\\ &\frac{d y}{d x}=\frac{1}{2}\\ &\text { Hence, } \mathrm{dy} / \mathrm{dx}=1 / 2 \end{aligned}

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