Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)+\tan ^{-1}\left(\frac{3+8 x}{3-3 x}\right)$, then $\frac{d y}{d x}=$
Options:
Solution:
2484 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{1+25 x^2}$
$\begin{aligned} & y=\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)+\tan ^{-1}\left(\frac{3+8 x}{8-3 x}\right)=\tan ^{-1}\left(\frac{5 x-x}{1+5 x \cdot x}\right)+\tan ^{-1}\left(\frac{\frac{3}{8}+x}{1-\frac{3}{8} \cdot x}\right) \\ & =\tan ^{-1}(5 x)-\tan ^{-1}(x)+\tan ^{-1}\left(\frac{3}{8}\right)+\tan ^{-1}(x) \\ & \Rightarrow y=\tan ^{-1}(5 x)+\tan ^{-1}\left(\frac{3}{8}\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{1+(5 x)^2} \times 5+0=\frac{5}{1+25 x^2}\end{aligned}$
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.