Question

If $a$ and $b$ are fixed non-zero constants, then the derivative of $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\operatorname{Cos} x$ is $ma+nb-p$ where

• A. $m=4 x^{3} ; n=\frac{-2}{x^{3}} ; p=\sin x$
• B. $m=\frac{-4}{x^{5}} ; n=\frac{2}{x^{3}} ; p=\sin x$
• C. $m=\frac{-4}{x^{5}} ; n=\frac{-2}{x^{3}} ; p=-\sin x$
• D. $m=4 x^{3} ; n=\frac{2}{x^{3}} ; p=-\sin x$

Answer 1

Answer: (B)

$\frac{ d }{ dx }\left(\frac{ a }{ x ^{4}}-\frac{ b }{ x ^{2}}+\cos x \right)=\left(-\frac{4 a }{ x ^{5}}+\frac{2 b }{ x ^{3}}-\sin x \right)$
$= ma + nb - p$
$m =-\frac{4}{ x ^{5}} ; n =\frac{+2}{ x ^{3}} ; p =\sin x$