#### Need solution for RD Sharma Maths Class 12 Chapter 11 Higher Order Derivatives Excercise Multiple Choice Questions Question 7

(c)

Hint:

We must know about the derivative of $\cos x$  and  $\sin x$ .

Given:

\begin{aligned} & f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots .(\cos n x+i \sin n x) \\ \end{aligned}  and  $f\left ( 1 \right )=1$ , then $f^{11}\left ( 1 \right )$  is equal to.

Explanation:

\begin{aligned} & f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots .(\cos n x+i \sin n x) \\ \end{aligned}

$\Rightarrow \quad \cos (x+2 x+3 x+\ldots .+n x)+i \sin (x+2 x+3 x+\ldots .+n x)$

Using Fourier series,

$\Rightarrow \quad \left[\cos \frac{n(n+1)}{2} x+i \sin \frac{n(n+1)}{2} x\right]$

Differentiate on both sides,

\begin{aligned} &f^{\prime}(x)=\frac{n(n+1)}{2}\left[-\sin \frac{n(n+1)}{2} x+i \cos \frac{n(n+1)}{2} x\right] \\ & \end{aligned}

$f^{\prime \prime}(x)=-\left(\frac{n(n+1)}{2}\right)^{2}\left[\cos \frac{n(n+1)}{2} x+i \sin \frac{n(n+1)}{2} x\right]$

\begin{aligned} f^{\prime \prime}(x) &=-\left(\frac{n(n+1)}{2}\right)^{2} f(x) \\ \end{aligned}

$\therefore f^{\prime \prime}(x) =-\left(\frac{n(n+1)}{2}\right)^{2} f(1)$

$=-\left(\frac{n(n+1)}{2}\right)^{2}$