Question

If $y={\tan }^{-1}\frac{4x}{1+5x^2}+{\tan }^{-1}\frac{2+3x}{3-2x}$ , then $\frac{dy}{dx}=$

• A. $\frac{1}{\left(\begin{array}{cc}1+& 25x^2\end{array}\right)}+\frac{2}{\left(\begin{array}{cc}1+& x^2\end{array}\right)}$
• B. $\frac{5}{\left(\begin{array}{cc}1+& 25x^2\end{array}\right)}+\frac{1}{\left(\begin{array}{cc}1+& x^2\end{array}\right)}$
• C. $\frac{5}{\left(\begin{array}{cc}1+& 25x^2\end{array}\right)}+\frac{1}{\left(\begin{array}{cc}1+& 25x^2\end{array}\right)}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

$$\begin{gathered} y={\tan }^{-1}\frac{4x}{1+5x^2}+{\tan }^{-1}\frac{2+3x}{3-2x} \\ ={\tan }^{-1}\left(\frac{5x-x}{1+5x^2}\right)+{\tan }^{-1}\left(\frac{{\displaystyle \frac{2}{3}}+x}{1-{\displaystyle \frac{2x}{3}}}\right) \\ ={\tan }^{-1}5x-{\tan }^{-1}x+{\tan }^{-1}\frac{2}{3}+{\tan }^{-1}x \\ ={\tan }^{-1}5x+{\tan }^{-1}\frac{2}{3} \end{gathered}$$

Now, differentiate with respect to $x$

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}5x\right)+\frac{d}{dx}\left({\tan }^{-1}\frac{2}{3}\right) \\ =\left(\frac{1}{1+25x^2}\right)\times 5+0 \\ =\frac{5}{1+25x^2} \end{gathered}$$

So, $\frac{dy}{dx}=\frac{5}{1+25x^2}$

Hence, Option (D) is the correct answer.