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