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If $y=\tan ^{-1} \frac{4 x}{1+5 x^{2}}+\tan ^{-1} \frac{2+3 x}{3-2 x}$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{5}{1+25 x^{2}}$
$$
\begin{aligned}
y &=\tan ^{-1} \frac{4 x}{1+5 x^{2}}+\tan ^{-1} \frac{2+3 x}{3-2 x} \\
&=\tan ^{-1} \frac{5 x-x}{1+5 x \cdot x}+\tan ^{-1} \frac{\frac{2}{3}+x}{1-\frac{2}{3} \cdot x} \\
&=\tan ^{-1} 5 x-\tan ^{-1} x+\tan ^{-1} \frac{2}{3}+\tan ^{-1} x \\
\Rightarrow & \frac{d y}{d x}=\frac{5}{1+25 x^{2}}
\end{aligned}
$$
\begin{aligned}
y &=\tan ^{-1} \frac{4 x}{1+5 x^{2}}+\tan ^{-1} \frac{2+3 x}{3-2 x} \\
&=\tan ^{-1} \frac{5 x-x}{1+5 x \cdot x}+\tan ^{-1} \frac{\frac{2}{3}+x}{1-\frac{2}{3} \cdot x} \\
&=\tan ^{-1} 5 x-\tan ^{-1} x+\tan ^{-1} \frac{2}{3}+\tan ^{-1} x \\
\Rightarrow & \frac{d y}{d x}=\frac{5}{1+25 x^{2}}
\end{aligned}
$$
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