{d over {dx}}left( {{1 over {{x^4}sec x}}} ri | vedclass

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Question

(b) $\frac{d}{{dx}}\left( {\frac{1}{{{x^4}\sec x}}} \right) = \frac{d}{{dx}}\left( {\frac{{\cos x}}{{{x^4}}}} \right)$$ = \frac{{{x^4}( - \sin x) - \c

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $\frac{d}{{dx}}\left( {\frac{1}{{{x^4}\sec x}}} \right) = \frac{d}{{dx}}\left( {\frac{{\cos x}}{{{x^4}}}} \right)$$ = \frac{{{x^4}( - \sin x) - \cos x(4{x^3})}}{{{{({x^4})}^2}}}$
$ = \frac{{ - {x^3}(x\sin x + 4\cos x)}}{{{x^8}}} = \frac{{ - (x\sin x + 4\cos x)}}{{{x^5}}}$.

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

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