How do you differentiate  f(x)=e^sqrt(1/x^2-x) using the chain rule.?

Aug 11, 2018

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

Explanation:

Given,
$f \left(x\right) = {e}^{\sqrt{\frac{1}{x} ^ 2 - x}}$
Let
$y = f \left(x\right)$
Then,
$y = {e}^{\sqrt{\frac{1}{x} ^ 2 - x}}$
Let,
$u = \sqrt{\frac{1}{x} - {x}^{2}}$
Then,
$y = {e}^{u}$
Differentiating both sides, wrt x

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} . \frac{\mathrm{du}}{\mathrm{dx}}$

Now,
$u = \sqrt{\frac{1}{x} - {x}^{2}}$

Let,
$v = \frac{1}{x} - {x}^{2}$
Then,
$u = \sqrt{v}$
Differentiating both sides, wrt x

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{v}} \frac{\mathrm{dv}}{\mathrm{dx}}$

Thus,

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \frac{1}{2 \sqrt{v}} \frac{\mathrm{dv}}{\mathrm{dx}}$
Again,

$v = \frac{1}{x} - {x}^{2}$
Let
$r = \frac{1}{x}$
Differentiating wrt x
$\frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{x} ^ 2$
$s = {x}^{2}$
Differentiating wrt x
$\frac{\mathrm{ds}}{\mathrm{dx}} = 2 x$
now,
$v = r - s$

Differentiating wrt x
$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dr}}{\mathrm{dx}} - \frac{\mathrm{ds}}{\mathrm{dx}}$
Thus,
$\frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{x} ^ 2 - 2 x$
Substituting for  v & (dv)/(dx) in $\frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{v}} \frac{\mathrm{dv}}{\mathrm{dx}}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\frac{1}{x} - {x}^{2}}} \left(- \frac{1}{x} ^ 2 - 2 x\right)$

Substituting for u & (du)/(dx) in $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} . \frac{\mathrm{du}}{\mathrm{dx}}$

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

Aug 11, 2018

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left(\frac{2}{x} ^ 3 + 1\right) {e}^{\sqrt{\frac{1}{x} ^ 2 - x}} / \sqrt{\frac{1}{x} ^ 2 - x}$

Explanation:

Here ,

$f \left(x\right) = y = {e}^{\sqrt{\frac{1}{x} ^ 2 - x}}$

Let ,

$y = {e}^{u}$ , $u = \sqrt{v} \mathmr{and} v = \frac{1}{x} ^ 2 - x = {x}^{-} 2 - x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u}$ , $\frac{\mathrm{du}}{\mathrm{dv}} = \frac{1}{2 \sqrt{v}} \mathmr{and} \frac{\mathrm{dv}}{\mathrm{dx}} = - 2 {x}^{-} 3 - 1 = \frac{- 2}{x} ^ 3 - 1$

Using Chain Rule:

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dv)(dv)/(dx)

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \frac{1}{2 \sqrt{v}} \left(- \frac{2}{x} ^ 3 - 1\right)$

Subst. back $u = \sqrt{v}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\sqrt{v}} \frac{1}{2 \sqrt{v}} \left(- \frac{2}{x} ^ 3 - 1\right)$

Now ,subst. $v = \frac{1}{x} ^ 2 - x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\sqrt{\frac{1}{x} ^ 2 - x}} \frac{1}{2 \sqrt{\frac{1}{x} ^ 2 - x}} \left(- \frac{2}{x} ^ 3 - 1\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left(\frac{2}{x} ^ 3 + 1\right) {e}^{\sqrt{\frac{1}{x} ^ 2 - x}} / \sqrt{\frac{1}{x} ^ 2 - x}$