Question
Differentiate the following w. r. t. x.$\left(\frac{\left(x^2+3\right)^2 \sqrt[3]{\left(x^3+5\right)^2}}{\sqrt{\left(2 x^2+1\right)^3}}\right)$
✓
Answer
Let $\begin{aligned} y & =\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+x-2\right)}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+4 x-3 x-2\right)}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-[2 x(3 x+2)-(3 x+2)]}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-1)}\right)=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)\end{aligned}$$
y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
\therefore \quad \frac{d y}{d x} & =\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2}
\end{aligned}
$
y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
\therefore \quad \frac{d y}{d x} & =\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2}
\end{aligned}
$
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