Differentiate the following functions with respect to x: ^ -1 ( x… - Vidyadip

Question

Differentiate the following functions with respect to x:
$\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$

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Answer

Let $\text{y}=\tan^{-1}\Big(\frac{\text{x}}{1+6\text{x}^3}\Big)$
$\Rightarrow\ \text{y}=\tan^{-1}\Big(\frac{3\text{x}-2\text{x}}{1+(3\text{x})(2\text{x})}\Big)$
$\Rightarrow\text{y}=\tan^{-1}3\text{x}-\tan^{-1}2\text{x}$
$\Big[\text{Since}, \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
Differentiate it with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+(3\text{x})^2}\frac{\text{d}}{\text{dx}}(3\text{x})-\frac{1}{1+(2\text{x})^3}\frac{\text{d}}{\text{dx}}(2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+9\text{x}^2}(3)-\frac{1}{1+4\text{x}^2}(2)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3}{1+9\text{x}^2}-\frac{2}{1+4\text{x}^2}$

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