If y = operatorname{Tan}^{-1}left(frac{3x - x | vedclass

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(D) Given $y = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$.Using the for

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(D) Given $y = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$.Using the formula $\operatorname{Tan}^{-1}(A) + \operatorname{Tan}^{-1}(B) = \operatorname{Tan}^{-1}\left(\frac{A+B}{1-AB}\right)$,we can simplify the expression.Let $u = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ and $v = \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$.Using the derivative formula $\frac{d}{dx}(\operatorname{Tan}^{-1}(f(x))) = \frac{f'(x)}{1 + (f(x))^2}$.For $u = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$,at $x=0$,$\frac{du}{dx} = \frac{d}{dx}\left(\frac{3x - x^3}{1 - 3x^2}\right) \Big|_{x=0} = \frac{(3 - 3x^2)(1 - 3x^2) - (3x - x^3)(-6x)}{(1 - 3x^2)^2} \Big|_{x=0} = \frac{(3)(1) - 0}{1^2} = 3$.For $v = \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$,at $x=0$,$\frac{dv}{dx} = \frac{d}{dx}\left(\frac{7x}{1 - 12x^2}\right) \Big|_{x=0} = \frac{7(1 - 12x^2) - 7x(-24x)}{(1 - 12x^2)^2} \Big|_{x=0} = \frac{7(1) - 0}{1^2} = 7$.Thus,$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = 3 + 7 = 10$.

If $y = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$,then at $x = 0$,$\frac{dy}{dx} = $

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If $y = \operatorname{Tan}^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{7x}{1 - 12x^2}\right)$,then at $x = 0$,$\frac{dy}{dx} = $

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