Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\left(1+x^2\right) \tan ^{-1} x-x$, then $\frac{d y}{d x}$ is
Options:
Solution:
2221 Upvotes
Verified Answer
The correct answer is:
$2 x \tan ^{-1} x$
Given,
$$
y=\left(1+x^2\right) \tan ^{-1} x-x
$$
Differentiating the given function w.r.t. $x$
$$
\begin{aligned}
\frac{d y}{d x} & =\left(1+x^2\right) \frac{d}{d x}\left(\tan ^{-1} x\right)+\tan ^{-1} x \frac{d}{d x}\left(1+x^2\right)-\frac{d}{d x}(x) \\
& =\left(1+x^2\right) \frac{1}{\left(1+x^2\right)}+\tan ^{-1} x(2 x)-1 \\
& =1+2 x \tan ^{-1} x-1=2 x \tan ^{-1} x
\end{aligned}
$$
$$
y=\left(1+x^2\right) \tan ^{-1} x-x
$$
Differentiating the given function w.r.t. $x$
$$
\begin{aligned}
\frac{d y}{d x} & =\left(1+x^2\right) \frac{d}{d x}\left(\tan ^{-1} x\right)+\tan ^{-1} x \frac{d}{d x}\left(1+x^2\right)-\frac{d}{d x}(x) \\
& =\left(1+x^2\right) \frac{1}{\left(1+x^2\right)}+\tan ^{-1} x(2 x)-1 \\
& =1+2 x \tan ^{-1} x-1=2 x \tan ^{-1} x
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.