How do you differentiate  f(x)=sqrt(e^((lnx-2)^2  using the chain rule.?

Mar 25, 2018

f^'(x)=[(lnx-2)*e^((lnx-2)^2)]/(x*sqrt(e^((lnx-2)^2)))=1/x(lnx-2)sqrt(e^((lnx-2)^2)

Explanation:

Here,
f(x)=y=sqrt(e^((lnx-2)^2)

Let us substitute,

$y = \sqrt{u} , u = {e}^{w} \mathmr{and} w = {\left(\ln x - 2\right)}^{2}$

$\implies \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}} , \frac{\mathrm{du}}{\mathrm{dw}} = {e}^{w} \mathmr{and} \frac{\mathrm{dw}}{\mathrm{dx}} = 2 {\left(\ln x - 2\right)}^{1} \cdot \frac{1}{x}$

Using Chain Rule ,

color(red)((dy)/(dx)=(dy)/(du)xx(du)/(dw)xx(dw)/(dx)

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cancel{2} \sqrt{u}} \times {e}^{w} \times \frac{\cancel{2} \left(\ln x - 2\right)}{x}$

Substituting value of $u , v \mathmr{and} w$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{{e}^{w}}} \times {e}^{{\left(\ln x - 2\right)}^{2}} \times \frac{\ln x - 2}{x}$

(dy)/(dx)=[(lnx-2)*e^((lnx-2)^2)]/(x*sqrt(e^((lnx-2)^2)

=>f^'(x)=1/x(lnx-2)sqrt(e^((lnx-2)^2)