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MHT CET · Maths · Differentiation

If \(y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)

  1. A \(\frac{1}{1+25 x^2}\)
  2. B \(\frac{5}{1+25 x^2}\)
  3. C \(\frac{1}{1+5 x^2}\)
  4. D \(\frac{5}{1+5 x^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{5}{1+25 x^2}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right) \\ & =\tan ^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3} x}\right)+\tan ^{-1}\left(\frac{5 x-x}{1+5 x^2}\right) \\ & =\tan ^{-1} \frac{2}{3}+\tan ^{-1} x+\tan ^{-1} 5 x-\tan ^{-1} x \\ \therefore \quad & y=\tan ^{-1} \frac{2}{3}+\tan ^{-1} 5 x \\ \therefore \quad & \frac{d y}{d x}=\frac{1}{1+(5 x)^2} \cdot 5=\frac{5}{1+25 x^2}\end{aligned}\)