If y=_k=1^6k^-1(35 kx-45 kx), then dydx at x=0 is _______.

Question

If $y=\displaystyle\sum_{k=1}^6k\cos^{-1}\left(\dfrac35\cos kx-\dfrac45\sin kx\right)$, then $\dfrac{dy}{dx}$ at $x=0$ is \_\_\_\_\_\_\_.

Answer: 91

Accepted Answer

Let $\cos a=\dfrac35,\ \sin a=\dfrac45$. Then $\cos a\cos kx-\sin a\sin kx=\cos(kx+a)$, so $y=\sum_{k=1}^6k(kx+a)=\sum_{k=1}^6(k^2x+ak)$.

$\dfrac{dy}{dx}=\sum_{k=1}^6k^2=\dfrac{6\cdot7\cdot13}{6}=91$.

Solutions · 1