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KCET · Maths · Differentiation

If \(a\) and \(b\) are fixed non-zero constants, then the derivative of \(\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x\) is \(m a+n b-p\), where

  1. A \(m=4 x^{3}, n=\frac{-2}{x^{3}}\) and \(p=\sin x\)
  2. B \(m=\frac{-4}{x^{5}}, n=\frac{2}{x^{3}}\) and \(p=\sin x\)
  3. C \(m=\frac{-4}{x^{5}}, n=\frac{-2}{x^{3}}\) and \(p=\sin x\)
  4. D \(m=4 x^{3}, n=\frac{2}{x^{3}}\) and \(p=-\sin x\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(m=\frac{-4}{x^{5}}, n=\frac{2}{x^{3}}\) and \(p=\sin x\)

Step-by-step Solution

Detailed explanation

Given, function is \(\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x\).
Let \(f(x)=\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x\)
Differentiating w.r.t. \(x\),
\(\begin{gathered}
\frac{d}{d x} f(x)=\frac{d}{d x}\left(\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x\right) \\
=-\frac{4 a}{x^{5}}+\frac{2 b}{x^{3}}-\sin x \\
=m a+n b-P \\
m=-\frac{4}{x^{5}}, n=\frac{2}{x^{3}} \text { and } p=\sin x
\end{gathered}\)