# Find dy/dx if?

## ${x}^{\cos} y + {y}^{\sin} x = 1$

Apr 13, 2018

(dy)/(dx)=-(y^sinxcosxlny+x^cosycosy/x)/(y^sinxsinx/y- x^cosysinylnx)

#### Explanation:

Here,

${x}^{\cos} y + {y}^{\sin} x = 1$

Let, $u + v = 1 , w h e r e ,$

$u = {x}^{\cos} y \mathmr{and} v = {y}^{\sin} x$

Taking log both sides,

$\implies \ln u = \ln {x}^{\cos} y \mathmr{and} \ln v = \ln {y}^{\sin} x$

$\implies \ln u = \cos y \ln x \ldots \to \left(1\right) \mathmr{and} \ln v = \sin x \ln y \ldots \to \left(2\right)$

Diff$. \to \left(1\right) w . r . t . x$ "using "color(blue)"Product Rule and Chain Rule"

$\frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}} = \cos y \times \frac{1}{x} + \ln x \left(- \sin y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = {x}^{\cos} y \left[\cos \frac{y}{x} - \sin y \ln x \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

Now diff$. \to \left(2\right) w . r . t . x$ "using "color(blue)"Product Rule and Chain Rule"

$\frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}} = \sin x \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \ln y \cos x$

$\frac{\mathrm{dv}}{\mathrm{dx}} = {y}^{\sin} x \left[\sin \frac{x}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \cos x \ln y\right]$

Hence, $u + v = 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} + \frac{\mathrm{dv}}{\mathrm{dx}} = 0 \implies \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{\mathrm{dv}}{\mathrm{dx}}$

${x}^{\cos} y \left[\cos \frac{y}{x} - \sin y \ln x \frac{\mathrm{dy}}{\mathrm{dx}}\right] = - {y}^{\sin} x \left[\sin \frac{x}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \cos x \ln y\right]$

{y^sinxsinx/y-x^cosysinylnx}(dy)/(dx)=-y^sinxcosxlny- x^cosycosy/x

(dy)/(dx)=-(y^sinxcosxlny+x^cosycosy/x)/(y^sinxsinx/y- x^cosysinylnx)

Apr 13, 2018

dy/dx=-(cos(y)x^((cosy)-1)+(lny)y^sinxcosx)/((sinx)y^((sinx)-1)-x^cosy(siny)lnx

#### Explanation:

if $u , v$ are two functions of $x$
then $\frac{d}{\mathrm{dx}} {u}^{v} = {u}^{v} \cdot \ln \left(u\right) \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot {u}^{v - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\textcolor{red}{\text{an easy way to memorize this rule is}}$$\textcolor{g r e e n}{\text{ to consider the function u to be a constant and differentiate it}}$ $\textcolor{b l u e}{\text{ then consider v to be a constant and differentiate it}}$
to get both the first term and second term

so consider ${x}^{\cos y} = {z}_{1}$
and $\left({y}^{\sin} x\right) = {z}_{2}$
so the upper formula will be ${z}_{1} + {z}_{2} = 1$

differentiate

${z}_{1} ' + {z}_{2} ' = 0$

${z}_{1} ' = \left(- \sin y\right) \frac{\mathrm{dy}}{\mathrm{dx}} \left(\ln x\right) {x}^{\cos} \left(y\right) + \cos \left(y\right) \cdot {x}^{\left(\cos y\right) - 1} \cdot 1$

${z}_{2} ' = \left(\ln y\right) {y}^{\sin} x \cos x + \left(\sin x\right) {y}^{\left(\sin x\right) - 1} \frac{\mathrm{dy}}{\mathrm{dx}}$

substitute for ${z}_{1} ' , {z}_{2} '$

$\left(- \sin y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \left(\ln x\right) \cdot {x}^{\cos} y + \cos \left(y\right) {x}^{\left(\cos y\right) - 1} + \left(\ln y\right) {y}^{\sin} x \cos x + \left(\sin x\right) {y}^{\left(\sin x\right) - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

simplify
dy/dx=-((cos(y)x^((cosy)-1)+(lny)y^sinxcosx))/((sinx)y^((sinx)-1)-x^cosy(siny)lnx