If f(x) = sum_{p=1}^7 p^2 sin^{-1}left(frac{4 | vedclass

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(D) Given $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$.Let $\cos \alpha = \frac{4}{5}$ and $\sin \alpha

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$784$

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(D) Given $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$.Let $\cos \alpha = \frac{4}{5}$ and $\sin \alpha = \frac{3}{5}$.Then the expression becomes $\sin^{-1}(\sin(px) \cos \alpha - \cos(px) \sin \alpha) = \sin^{-1}(\sin(px - \alpha)) = px - \alpha$.Thus, $f(x) = \sum_{p=1}^7 p^2(px - \alpha) = \sum_{p=1}^7 (p^3 x - p^2 \alpha) = x \sum_{p=1}^7 p^3 - \alpha \sum_{p=1}^7 p^2$.Taking the derivative with respect to $x$:$\frac{df}{dx} = \frac{d}{dx} \left( x \sum_{p=1}^7 p^3 - \alpha \sum_{p=1}^7 p^2 \right) = \sum_{p=1}^7 p^3$.The sum of cubes is given by $\left(\frac{n(n+1)}{2}\right)^2$ for $n=7$.$\frac{df}{dx} = \left(\frac{7(8)}{2}\right)^2 = (28)^2 = 784$.

If $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$, then the value of $\frac{df}{dx}$ at $x = 1$ is (Given that $\sin^{-1}(\sin x) = x$)

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