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Question

Mathematics Question on Differentiability

Derivative of log(secθ+tanθ)log \left(sec\,\theta +tan \,\theta\right) with respect to secθsec\, \theta at θ=π/4\theta = \pi/4 is

A

00

B

11

C

12\frac{1}{\sqrt2}

D

2\sqrt2

Answer

11

Explanation

Solution

Let u=log(secθ+tanθ)u=log (sec\, \theta+tan \, \theta) and v=secθv=sec\, \theta
On differentiating both sides w.r.t. θ,\theta, we get
dudθ=1(secθ+tanθ)(secθtanθ+sec2θ)\frac{du}{d \theta}=\frac{1}{(\sec \theta+\tan \theta)}\left(\sec \theta \tan \theta+\sec ^{2} \theta\right)
and dvdθ=secθtanθ\frac{dv}{d \theta}=\sec \theta \tan \theta
dudv=dudθdvdθ\therefore \frac{du}{dv}=\frac{\frac{du}{d \theta}}{\frac{dv}{d \theta}}
=secθ(tanθ+secθ)(secθ+tanθ)×secθtanθ=cotθ=\frac{\sec \theta(\tan \theta+\sec \theta)}{(\sec \theta+\tan \theta) \times \sec \theta \tan \theta}=\cot \theta
dudv(θ=π4)=cotπ4=1\Rightarrow \frac{du}{dv\left(\theta=\frac{\pi}{4}\right)}=\cot \frac{\pi}{4}=1