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3-1.Vectors
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${d \over {dx}}\left( {{1 \over {{x^4}\sec x}}} \right) = $
A${{x\sin x + 4\cos x} \over {{x^5}}}$
B${{ - (x\sin x + 4\cos x)} \over {{x^5}}}$
C${{4\cos x - x\sin x} \over {{x^5}}}$
DNone of these
Solution
(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}}}$.
$ = \frac{{ – {x^3}(x\sin x + 4\cos x)}}{{{x^8}}} = \frac{{ – (x\sin x + 4\cos x)}}{{{x^5}}}$.
Standard 11
Physics
