d dx ( 1 x^4 x ) = - Vidyadip

MCQ

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

A
${{x\sin x + 4\cos x} \over {{x^5}}}$
✓
${{ - (x\sin x + 4\cos x)} \over {{x^5}}}$
C
${{4\cos x - x\sin x} \over {{x^5}}}$
D
None of these

✓

Answer

Correct option: B.

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

b
(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}}}$.

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