If $$y = \sum_{k=1}^{6} k\cos^{-1}\left\{\frac{3}{5}\cos kx - \frac{4}{5}\sin kx\right\}$$ then $$\frac{dy}{dx}$$ at $$x = 0$$ is ___________.

We have to differentiate the function

$$y=\sum_{k=1}^{6}k\,\cos^{-1}\!\Bigg\{\frac35\cos kx-\frac45\sin kx\Bigg\}$$

and then evaluate $$\dfrac{dy}{dx}$$ at $$x=0$$.

First we inspect the expression inside the inverse cosine. Write

$$A_k(x)=\frac35\cos kx-\frac45\sin kx.$$

Notice that the numerical coefficients satisfy

$$\left(\frac35\right)^{\!2}+\left(\frac45\right)^{\!2}=\frac{9}{25}+\frac{16}{25}=1.$$

Because the sum of squares is $$1,$$ the pair $$\left(\frac35,-\frac45\right)$$ can be interpreted as $$\bigl(\cos\varphi,-\sin\varphi\bigr)$$ for some angle $$\varphi$$ lying in the first quadrant. Indeed, choose $$\varphi$$ such that

$$\cos\varphi=\frac35,\qquad\sin\varphi=\frac45.$$

Now use the standard identity

$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Putting $$\alpha=kx$$ and $$\beta=\varphi,$$ we get

$$\cos(kx+\varphi)=\cos kx\cos\varphi-\sin kx\sin\varphi =\frac35\cos kx-\frac45\sin kx =A_k(x).$$

So we have shown

$$A_k(x)=\cos(kx+\varphi)\quad\text{where}\quad\cos\varphi=\frac35,\ \sin\varphi=\frac45.$$

Hence the given sum becomes

$$y=\sum_{k=1}^{6}k\,\cos^{-1}\!\Bigl(\cos(kx+\varphi)\Bigr).$$

For the principal value of $$\cos^{-1}(\cdot),$$ the range is $$[0,\pi].$$ The fixed angle $$\varphi$$ equals $$\cos^{-1}\!\frac35\approx0.927\,\text{rad},$$ which lies strictly between $$0$$ and $$\pi.$$ When $$x$$ is very close to $$0$$, each argument $$kx+\varphi$$ also remains in the interval $$\bigl(0,\pi\bigr).$$ Therefore, for sufficiently small $$x$$ (including the point $$x=0$$ itself) we may legitimately use

$$\cos^{-1}\bigl(\cos\theta\bigr)=\theta\quad\text{whenever}\quad 0\le\theta\le\pi.$$

Applying this to every term, we get in a neighbourhood of $$x=0$$

$$\cos^{-1}\!\Bigl(\cos(kx+\varphi)\Bigr)=kx+\varphi.$$

Consequently,

$$y=\sum_{k=1}^{6}k\,(kx+\varphi)=\sum_{k=1}^{6}\bigl(k^{2}x+k\varphi\bigr) =\Bigl(\sum_{k=1}^{6}k^{2}\Bigr)x+\Bigl(\sum_{k=1}^{6}k\Bigr)\varphi.$$

Differentiate term by term with respect to $$x$$. The constant parts involving $$\varphi$$ vanish, leaving

$$\frac{dy}{dx}=\sum_{k=1}^{6}k^{2}.$$

Now evaluate the simple finite sum of squares:

$$\sum_{k=1}^{6}k^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2} =1+4+9+16+25+36=91.$$

Finally, substitute $$x=0$$ (though the derivative is already independent of $$x$$ in the chosen neighbourhood):

$$\left.\frac{dy}{dx}\right|_{x=0}=91.$$

Hence, the correct answer is Option A, $$91$$.