Find Dy/dx Help i have no idea how to use the chain rule here? y=ln(tan^-1(ex))

Apr 19, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{e}{\left(1 + {e}^{2} {x}^{2}\right) {\tan}^{-} 1 \left(e x\right)}$

Explanation:

We know that,

color(red)((1) d/(dX)(lnX)=1/X

color(red)((2) d/(dX)(tan^-1X)=1/(1+X^2

Here,

$y = \ln \left({\tan}^{-} 1 \left(e x\right)\right)$

Take, $y = \ln u \mathmr{and} u = {\tan}^{-} 1 \left(e x\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u} \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {\left(e x\right)}^{2}} \frac{d}{\mathrm{dx}} \left(e x\right)$

where,u=tan^-1(ex) and color(orange)(d/(dx)(ex)=e

$i . e . \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{{\tan}^{-} 1 \left(e x\right)} \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{e}{1 + {e}^{2} {x}^{2}}$

$\text{Using "color(blue)"Chain Rule:}$

color(blue)((dy)/(dx)=(dy)/(du)xx(du)/(dx)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\tan}^{-} 1 \left(e x\right)} \times \frac{e}{1 + {e}^{2} {x}^{2}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{e}{\left(1 + {e}^{2} {x}^{2}\right) {\tan}^{-} 1 \left(e x\right)}$

Note:

If y=ln(tan^-1(e^x)). then ,take,v=e^x=>(dv)/(dx)=color(orange)(d/(dx)(e^x)=e^x.

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} / \left(\left(1 + {\left({e}^{x}\right)}^{2}\right) {\tan}^{-} 1 \left({e}^{x}\right)\right)$