Question:

<b>Differentiate Sum of Inverse Trig</b><br> $\dfrac{3}{5}\cos kx-\dfrac{4}{5}\sin kx = \cos\!\left(kx+\phi\right)$ where $\cos\phi=3/5$, $\sin\phi=4/5$, $\phi=\sin^{-1}(4/5)$.<br> So $\cos^{-1}\!\left(\dfrac{3}{5}\cos kx-\dfrac{4}{5}\sin kx\right)=\cos^{-1}(\cos(kx+\phi))=kx+\phi$ (for appropriate range).<br> $y=\sum_{k=1}^6 k(kx+\phi)=\left(\sum_{k=1}^6 k^2\right)x+\phi\sum_{k=1}^6 k$.<br> $\dfrac{dy}{dx}=\sum_{k=1}^6 k^2 = 1+4+9+16+25+36=91$.<br> <b>Answer: 91</b><br> <b>Key concept:</b> Write $a\cos\theta+b\sin\theta$ in the form $R\cos(\theta+\phi)$; then $\cos^{-1}(\cos(\theta))=\theta$ in the right range.<br> <b>Trap:</b> Not recognising the $R\cos(\theta+\phi)$ form; trying to differentiate the sum directly term by term.