MHT CET · Maths · Differentiation
The derivative of \(\cot ^{-1} x\) w.r.t \(\log \left(1+x^{2}\right)\) is
- A \(-2 x\)
- B \(-\frac{1}{2 x}\)
- C \(\frac{1}{2 x}\)
- D \(2 x\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2 x}\)
Step-by-step Solution
Detailed explanation
(C)
Let \(u=\cot ^{-1} x\) and \(v=\log \left(1+x^{2}\right)\)
\(\begin{array}{l}\frac{d u}{d x}=\frac{-1}{1+x^{2}} \text { and } \frac{d v}{d x}=\frac{2 x}{1+x^{2}} \\ \therefore \frac{d u}{d v}=\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}=\frac{\left(\frac{-1}{1+x^{2}}\right)}{\left(\frac{2 x}{1+x^{2}}\right)}=\frac{-1}{2 x}\end{array}\)
Let \(u=\cot ^{-1} x\) and \(v=\log \left(1+x^{2}\right)\)
\(\begin{array}{l}\frac{d u}{d x}=\frac{-1}{1+x^{2}} \text { and } \frac{d v}{d x}=\frac{2 x}{1+x^{2}} \\ \therefore \frac{d u}{d v}=\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}=\frac{\left(\frac{-1}{1+x^{2}}\right)}{\left(\frac{2 x}{1+x^{2}}\right)}=\frac{-1}{2 x}\end{array}\)
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