Question

If $y=\sum _{k=1}^{6}k{\cos }^{-1}\{\frac{3}{5}\cos kx-\frac{4}{5}\sin kx\}$, then $\frac{dy}{dx}atx=0$ is

Answer 1

Step1. Simplifying the term:

Given,

$y=\sum _{k=1}^{6}k{\cos }^{-1}\{\frac{3}{5}\cos kx-\frac{4}{5}\sin kx\}...\left(i\right)$

Let $$\begin{gathered} \cos \theta =\frac{3}{5} \\ \sin \theta =\frac{4}{5} \end{gathered}$$

Therefore, our equation becomes

$\begin{array}{rcl}& & k{\cos }^{-1}\{\cos \theta \cos kx-\sin \theta \sin kx\}\\ & =& k{\cos }^{-1}\left[\cos \right(kx+\theta \left)\right]\\ & =& k(kx+\theta )\\ & =& k^{2}x+k\theta \end{array}$

Step2. Finding differentiation:

Equation $\left(i\right)$ can be written as

$y=\sum _{k=1}^6[k^{2}x+k\theta ]$

Therefore,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d\sum _{k=1}^6[k^{2}x+k\theta ]}{dx}\\ & =& \sum _{k=1}^6[\frac{d(k^{2}x+k\theta )}{dx}]\\ & =& \sum _{k=1}^6[\frac{d\left(k^{2}x\right)}{dx}+\frac{d\left(k\theta \right)}{dx}]\\ & =& \sum _{k=1}^6[k^2\frac{dx}{dx}+0]\\ & =& \sum _{k=1}^6\left[k^2\right]\\ & =& 1^2+2^2+3^2+4^2+5^2+6^2\\ & =& \frac{6\times (6+1)\times (6\times 2+1)}{6}\left[\frac{\mathbf{n}\mathbf{\times }\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\right]\\ & =& 91\end{array}$

Hence, $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathit{a}\mathit{t}\mathbf{}\mathbf{}\mathit{x}\mathbf{=}\mathbf{0}$ is $\mathbf{91}$.