Question

If $f\left(x\right)=\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right)$, then the value of f '(0) is

• A. 1
• B. -1
• C. \N
• D. $\infty$

Answer 1

The correct option is A 1

Given that,
$f\left(x\right)=\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\mathrm{Differentiating}w.r.t.x\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\Rightarrow f\text{'}\left(x\right)=\frac{d}{\mathrm{dx}}\left(\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right\}\Rightarrow f\text{'}\left(x\right)=\frac{d\left(\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right\}}{d\left(\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right)}\times \frac{d\left(\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right)}{d\left(\frac{\pi }{4}+\frac{x}{2}\right)}\times \frac{d\left(\frac{\pi }{4}+\frac{x}{2}\right)}{\mathrm{dx}}\Rightarrow f\text{'}\left(x\right)=\frac{1}{\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)}\times {\sec }^2\left(\frac{\pi }{4}+\frac{x}{2}\right)\times \frac{1}{2}\Rightarrow f\text{'}\left(x\right)=\frac{\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)}{\sin \left(\frac{\pi }{4}+\frac{x}{2}\right)}\times \frac{1}{{\cos }^2\left(\frac{\pi }{4}+\frac{x}{2}\right)}\times \frac{1}{2}\Rightarrow f\text{'}\left(x\right)=\frac{1}{2\sin \left(\frac{\pi }{4}+\frac{x}{2}\right)\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)}\Rightarrow f\text{'}\left(x\right)=\frac{1}{\sin 2\left(\frac{\pi }{4}+\frac{x}{2}\right)}=\frac{1}{\sin \left(\frac{\pi }{2}+x\right)}=\frac{1}{\cos x}\Rightarrow f\text{'}\left(x\right)=\sec x\therefore f\text{'}\left(0\right)=\sec 0=1$