Question:
If $\mathrm{y}=\sum_{\mathrm{k}=1}^{6} \mathrm{k} \cos ^{-1}\left\{\frac{3}{5} \cos \mathrm{kx}-\frac{4}{5} \sin \mathrm{kx}\right\}$
then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=0$ is________.
Solution:
Put $\cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5} \quad 0<\alpha<\frac{\pi}{2}$
Now $\frac{3}{5} \cos \mathrm{kx}-\frac{4}{5} \sin \mathrm{kx}$
$=\cos \alpha \cdot \cos \mathrm{kx}-\sin \alpha \cdot \sin \mathrm{kx}$
$=\cos (\alpha+\mathrm{kx})$
As we have to find derivate at $x=0$
We have $\cos ^{-1}(\cos (\alpha+k x))$
$=(\alpha+\mathrm{kx})$
$\Rightarrow y=\sum_{k=1}^{6}(\alpha+k x)$
$\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{a \mathrm{x}=0}=\sum_{k=\mathrm{x}}^{6} \mathrm{k}=\frac{6 \times 7 \times 13}{6}=91$