Find the derivative: frac{d}{dx} left( frac{1 | vedclass

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(B) We need to find the derivative of $f(x) = \frac{1}{x^4 \sec x}$.Since $\frac{1}{\sec x} = \cos x$,we can rewrite the function as $f(x) = \frac{\co

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We need to find the derivative of $f(x) = \frac{1}{x^4 \sec x}$.Since $\frac{1}{\sec x} = \cos x$,we can rewrite the function as $f(x) = \frac{\cos x}{x^4}$.Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \cos x$ and $v = x^4$:$u' = -\sin x$ and $v' = 4x^3$.Substituting these into the quotient rule:$\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}$.Thus,the correct option is $B$.

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

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