frac{d}{dx}left( frac{1}{x^4 sec x} right) = | vedclass

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(B) We know that $\frac{1}{\sec x} = \cos x$. Therefore,the expression becomes $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right)$. Using the quotient rul

Vedclass · Accepted Answer

$\frac{-(x \sin x + 4 \cos x)}{x^5}$

Vedclass · Answer

(B) We know that $\frac{1}{\sec x} = \cos x$. Therefore,the expression becomes $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right)$. Using the quotient rule $\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \cos x$ and $v = x^4$: $\frac{d}{dx}\left( \frac{\cos x}{x^4} \right) = \frac{x^4(-\sin x) - \cos x(4x^3)}{(x^4)^2}$ $= \frac{-x^4 \sin x - 4x^3 \cos x}{x^8}$ $= \frac{-x^3(x \sin x + 4 \cos x)}{x^8}$ $= \frac{-(x \sin x + 4 \cos x)}{x^5}$.

$\frac{d}{dx}\left( \frac{1}{x^4 \sec x} \right) = $

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