# Find y' in these 2 equations?

## $x . {e}^{- \frac{y}{2}}$$+ y . {e}^{- \frac{x}{2}}$$= \sin \left({x}^{2} + {y}^{2}\right)$ $\mathmr{and}$ $y = {x}^{\ln x} . {\left(\sec x\right)}^{3 x}$

Aug 12, 2018

$y ' = \frac{4 x \cos \left({x}^{2} + {y}^{2}\right) - 2 {e}^{- \frac{y}{2}} + y {e}^{- \frac{x}{2}}}{2 {e}^{- \frac{x}{2}} - x {e}^{- \frac{y}{2}} - 4 y \cos \left({x}^{2} + {y}^{2}\right)}$

#### Explanation:

Let , $y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = {y}_{1}$

Here ,

$\left(1\right) x \cdot {e}^{- \frac{y}{2}} + y \cdot {e}^{- \frac{x}{2}} = \sin \left({x}^{2} + {y}^{2}\right)$

Diff. w.r.t. $x$,using product and chain rules :

$x {e}^{- \frac{y}{2}} \left(- \frac{1}{2}\right) {y}_{1} + {e}^{- \frac{y}{2}} + y {e}^{- \frac{x}{2}} \left(- \frac{1}{2}\right) + {e}^{- \frac{x}{2}} \cdot {y}_{1}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} = \cos \left({x}^{2} + {y}^{2}\right) \left[2 x + 2 y {y}_{1}\right]$

$\therefore x {e}^{- \frac{y}{2}} \left(- \frac{1}{2}\right) {y}_{1} + {e}^{- \frac{x}{2}} \cdot {y}_{1} - \cos \left({x}^{2} + {y}^{2}\right) 2 y {y}_{1}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} = 2 x \cos \left({x}^{2} + {y}^{2}\right) - {e}^{- \frac{y}{2}} - y {e}^{- \frac{x}{2}} \left(- \frac{1}{2}\right)$

${y}_{1} \left\{{e}^{- \frac{x}{2}} - \frac{x}{2} {e}^{- \frac{y}{2}} - 2 y \cos \left({x}^{2} + {y}^{2}\right)\right\}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} = 2 x \cos \left({x}^{2} + {y}^{2}\right) - {e}^{- \frac{y}{2}} + \frac{y}{2} {e}^{- \frac{x}{2}}$

${y}_{1} \left\{2 {e}^{- \frac{x}{2}} - x {e}^{- \frac{y}{2}} - 4 y \cos \left({x}^{2} + {y}^{2}\right)\right\}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} = 4 x \cos \left({x}^{2} + {y}^{2}\right) - 2 {e}^{- \frac{y}{2}} + y {e}^{- \frac{x}{2}}$

$\therefore {y}_{1} = \frac{4 x \cos \left({x}^{2} + {y}^{2}\right) - 2 {e}^{- \frac{y}{2}} + y {e}^{- \frac{x}{2}}}{2 {e}^{- \frac{x}{2}} - x {e}^{- \frac{y}{2}} - 4 y \cos \left({x}^{2} + {y}^{2}\right)}$

Aug 12, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2 \ln x - 1} \cdot \ln {x}^{2} + 3 {\left(\sec x\right)}^{6 x} \left[x \tan x + \ln \left(\sec x\right)\right]$

#### Explanation:

Let , $y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = {y}_{1}$

$\left(2\right) y = {x}^{\ln} x \cdot {\left(\sec x\right)}^{3 x} = u \cdot v \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(i\right)$

Let ,

$u = {x}^{\ln} x$ ,$\ldots \ldots \ldots \ldots \ldots \ldots \mathmr{and} v = {\left(\sec x\right)}^{3 x}$

Taking natural log both sides we get

$\ln u = \ln \left({x}^{\ln} x\right)$ , $\ldots \ldots \ldots \ldots \ldots . \mathmr{and} \ln v = \ln \left({\left(\sec x\right)}^{3 x}\right)$

$\therefore \ln u = \ln x \cdot \ln x = {\left(\ln x\right)}^{2} \ldots \ldots \ldots \ldots \mathmr{and} \ln v = 3 x \ln \left(\sec x\right)$

Diff.w.r.t. $x$

$\frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}} = 2 \ln x \cdot \frac{1}{x} \mathmr{and} \frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}}$=$3 x \cdot \frac{\sec x \tan x}{\sec} x + \ln \left(\sec x\right) \cdot 3$

$\frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \ln {x}^{2.} \ldots \ldots \ldots . . \mathmr{and} \frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}}$=$3 x \tan x + 3 \ln \left(\sec x\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{u}{x} \ln {x}^{2.} \ldots \ldots \ldots . . \mathmr{and} \frac{\mathrm{dv}}{\mathrm{dx}} = 3 v \left[x \tan x + \ln \left(\sec x\right)\right] \ldots . \left(i i\right)$

Now ,

$y = u \cdot v \to \text{Product Rule}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

From $\left(i\right) \mathmr{and} \left(i i\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\ln} x \left[\frac{u}{x} \ln {x}^{2}\right] + {\left(\sec x\right)}^{3 x} \left[3 v \left(x \tan x + \ln \left(\sec x\right)\right)\right]$

Subst. back values of $u \mathmr{and} v$
$\frac{\mathrm{dy}}{\mathrm{dx}}$=${x}^{\ln} x \left[{x}^{\ln} \frac{x}{x} \ln {x}^{2}\right] + {\left(\sec x\right)}^{3 x} \left[3 {\left(\sec x\right)}^{3 x} \left(x \tan x + \ln \left(\sec x\right)\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2 \ln x - 1} \cdot \ln {x}^{2} + 3 {\left(\sec x\right)}^{6 x} \left[x \tan x + \ln \left(\sec x\right)\right]$