Question

Find $\frac{dy}{dx}$ if $x^y+y^x=1$

Answer 1

Let $u=y^x$ and $v=x^y$
Now $u+v=1$

Differentiating w.r.t. $x$ ,

$\frac{d(u+v)}{dx}=\frac{d\left(1\right)}{dx}$

$\frac{d\left(u\right)}{dx}=\frac{d\left(v\right)}{dx}=0\cdots \left(i\right)$

Now, we will find the derivative of $u$ and $v$ separately.

Finding $\frac{d\left(u\right)}{dx}$

$u=y^x$
Taking $\text{log}$ ⁡ on both sides, we get

$\text{log}u=\text{log}(y{)}^x$
$\text{log}u=x\text{log}\left(y\right)$

as $\left(\text{log}\right(a^n)=n(\text{log}a)$

Differentiating both sides w.r.t. $x$, we get

$\frac{d\left(\text{log}u\right)}{dx}=\frac{d(x.\text{log}y)}{dx}$


$\frac{1}{u}\left(\frac{du}{dx}\right)=\frac{d(x.\text{log}y)}{dx}$

$\frac{1}{u}\frac{du}{dx}=\frac{d\left(x\text{log}y\right)}{dx}$

Using product rule : $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$

$\frac{1}{u}\frac{du}{dx}=\frac{dx}{dx}\text{log}y+\frac{d\left(\text{log}y\right)}{dx}.x$


$\frac{1}{u}\frac{du}{dx}=1.\text{log}y+x\cdot \frac{d\left(\text{log}y\right)}{dy}.\frac{dy}{dx}$

$\frac{1}{u}\frac{du}{dx}=\text{log}y+\frac{x}{y}.\frac{dy}{dx}$

$\frac{du}{dx}=u(\text{log}y+\frac{x}{y}.\frac{dy}{dx})$

Substituting $u=y^x$, we get

$\frac{du}{dx}=y^x(\text{log}y+\frac{x}{y}\frac{dy}{dx})$

$\frac{du}{dx}=y^x\text{log}y+y^{x-1}x.\frac{dy}{dx}\cdots \left(ii\right)$

Finding $\frac{d\left(v\right)}{dx}$

$v=x^y$

Taking $\text{log}$ both sides, we get,

$\text{log}v=\text{log}\left(x^y\right)$

$\text{log}v=y.\text{log}x$

(as $\text{log}\left(a^n\right)=n\text{log}a)$

Differentiating both sides w.r.t. $x$, we get

$\frac{d\left(\text{log}v\right)}{dx}=\frac{d(y.\log x)}{dx}$
$\frac{d\left(\text{log}v\right)}{dx}\left(\frac{dv}{dx}\right)=\frac{d(y.\text{log}x)}{dx}$

$\frac{1}{v}\times \frac{dv}{dx}=\frac{d(y.\text{log}x)}{dx}$

Using product rule , $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$

$\frac{1}{v}\left(\frac{dv}{dx}\right)=\frac{d\left(y\right)}{dx}.\text{log}x+\frac{d\left(\text{log}x\right)}{dx}.y$

$\frac{1}{v}\left(\frac{dv}{dx}\right)=\frac{dy}{dx}.\text{log}x+\frac{y}{x}$

$\frac{dv}{dx}=v(\frac{dy}{dx}\text{log}x+\frac{y}{x})$

Substituting $v=x^y$, we get

$\frac{dv}{dx}=x^y(\frac{dy}{dx}\text{log}x+\frac{y}{x})$

$\frac{dv}{dx}=x^y\text{log}x.\frac{dy}{dx}+x^y\times \frac{y}{x}\cdots \left(iii\right)$

From $\left(i\right)$

$\frac{d\left(u\right)}{dx}+\frac{d\left(v\right)}{dx}=0$

Substituting the value of (ii) and (iii) in (i), we get

$(y^x\text{log}y+y^{x-1}.x\frac{dy}{dx})+(x^y\text{log}x.\frac{dy}{dx}+x^y\frac{y}{x})=0$

$\frac{dy}{dx}(x^y\text{log}x+y^{x-1}.x)=-(x^{y-1}y+y^x\text{log}y)$

$\frac{dy}{dx}=\frac{-(x^{y-1}y+y^x\text{log}y)}{(x^y\text{log}x+y^{x-1}.x)}$