Question Number 68331 by Peculiar last updated on 08/Sep/19
![Differentiate y=ln tan^(−1) (3x^2 )_](https://www.tinkutara.com/question/Q68331.png)
$${Differentiate}\:{y}={ln}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \underset{} {\right)} \\ $$
Answered by MJS last updated on 08/Sep/19
![y=h(g(f(x))) y′=h′(g(f(x)))×g′(f(x))×f′(x) h(t)=ln t ⇒ h′(t)=(1/t) g(t)=arctan t ⇒ g′(t)=(1/(t^2 +1)) f(t)=3t^2 ⇒ f′(t)=6t ⇒ y′=((6x)/((9x^4 +1)arctan (3x^2 )))](https://www.tinkutara.com/question/Q68335.png)
$${y}={h}\left({g}\left({f}\left({x}\right)\right)\right) \\ $$$${y}'={h}'\left({g}\left({f}\left({x}\right)\right)\right)×{g}'\left({f}\left({x}\right)\right)×{f}'\left({x}\right) \\ $$$${h}\left({t}\right)=\mathrm{ln}\:{t}\:\Rightarrow\:{h}'\left({t}\right)=\frac{\mathrm{1}}{{t}} \\ $$$${g}\left({t}\right)=\mathrm{arctan}\:{t}\:\Rightarrow\:{g}'\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}\left({t}\right)=\mathrm{3}{t}^{\mathrm{2}} \:\Rightarrow\:{f}'\left({t}\right)=\mathrm{6}{t} \\ $$$$\Rightarrow \\ $$$${y}'=\frac{\mathrm{6}{x}}{\left(\mathrm{9}{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{arctan}\:\left(\mathrm{3}{x}^{\mathrm{2}} \right)} \\ $$
Answered by peter frank last updated on 09/Sep/19
![let u=tan^(−1) (3x^2 ) (du/dx)=((6x)/((1+9x^4 ))) y=ln u (dy/du)=(1/u)=(1/(tan^(−1) (3x^2 ))) by chain rule (dy/dx)=(dy/du).(du/dx) (dy/dx)=((6x)/(tan^(−1) (3x^2 ).(1+9x^4 )))](https://www.tinkutara.com/question/Q68340.png)
$${let}\:{u}=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \right) \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{6}{x}}{\left(\mathrm{1}+\mathrm{9}{x}^{\mathrm{4}} \right)} \\ $$$${y}=\mathrm{ln}\:{u} \\ $$$$\frac{{dy}}{{du}}=\frac{\mathrm{1}}{{u}}=\frac{\mathrm{1}}{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \right)} \\ $$$${by}\:{chain}\:{rule} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}.\frac{{du}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{6}{x}}{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}^{\mathrm{2}} \right).\left(\mathrm{1}+\mathrm{9}{x}^{\mathrm{4}} \right)} \\ $$$$ \\ $$$$ \\ $$