Question

If $y=\cos \frac{x}{2}\cdot \cos \frac{x}{2^2}\cdots \infty$, then $-\frac{y^{\prime }}{y}$ is

• A. $\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\cdots \infty$
• B. $\frac{1}{2}\cot \frac{x}{2}+\frac{1}{2^2}\cot \frac{x}{2^2}+\cdots \infty$
• C. $\frac{1}{2}\sec \frac{x}{2}+\frac{1}{2^2}\sec \frac{x}{2^2}+\cdots \infty$
• D. $\frac{1}{2^n}\sin \frac{x}{2}+\frac{1}{2^{n-1}}\sin \frac{x}{2^2}+\cdots \infty$

Answer 1

The correct option is A $\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\cdots \infty$

Taking $\log$ on both sides

$\log \left(y\right)$$=\log [\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{2^2}\right)....\infty ]$

$=\log \left(\cos \frac{x}{2}\right)+\log \left(\cos \frac{x}{2^2}\right)+....\infty$

Differentiating both sides w.r.t. $x$

$\frac{1}{y}\frac{dy}{dx}=\frac{-\sin \left(x/2\right)}{\cos \left(x/2\right)}\frac{1}{2}+\frac{-\sin \left(x/2^2\right)}{\cos \left(x/2^2\right)}\frac{1}{2^2}+....\infty$

Hence correct answer is:

$\frac{-y^{\prime }}{y}=\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+....\infty$