\(y=\log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) ; \)
\(\frac{d y}{d x}=\frac{d}{d x} \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\)
\(\frac{d y}{d x}=\frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}\left[\sec ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)\right] \cdot \frac{1}{2}\)
\(=\frac{1}{2} \cdot \frac{1}{\frac {sec \left(\frac {\pi}{4}+\frac{x}{2}\right)}{\cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}} \cdot \frac{1}{\cos 2\left(\frac{\pi}{4}+\frac{x}{2}\right)} \)
\(\frac{d y}{d x}=\frac{1}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)\)
\(\frac {dy}{dx} = \frac {1}{sin 2 \left(\frac {\pi}{4}+ \frac {x}{2}\right)}\)
\(\Rightarrow \frac {dy}{dx} = \frac {1}{sin \left(\frac {\pi}{2} + x\right)}\)
\(\frac {dy}{dx} = \frac {1}{cos \,x} = sec \,x\)
प्रश्नानुसार, बायाँ पक्ष \(\frac {dy}{dx}\) = sec x ; sec x - sec x = 0 Proved.
