Question

Find $\frac{dy}{dx}$, if $y^x+x^y+x^x=a^b$.

Answer 1

Simplification of given data

Let $u=y^x,v=x^y\text{and}w=x^x$

Now, $u+v+w=a^b$

Differentiating w.r.t. $x$

$\frac{d(u+v+w)}{dx}=\frac{d\left(a^b\right)}{dx}$

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

(as $a^b$ is constant)

Now, we will find the derivative of $u,v\text{and}w$ separately.

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

$u=y^x$

Taking $\log$ both sides,
we get,
$\log u=\log \left(y^x\right)$

$\log u=x.\log y$ $\left(\log \right(a^n)=n\log a)$

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

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

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

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

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

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

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

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

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

Substituing $u=y^x$, we get,

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

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

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

$v=x^y$

Taking $\log$ both sides,
we get,

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

$\log v=y.\log x$ $\left(\log \right(a^n)=n\log a)$

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

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

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

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

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

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

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

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

Substituing $v=x^y$, we get,

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

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

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

$w=x^x$

Taking $\log$ both sides, we get,

$\log w=\log \left(x^x\right)$

$\log w=x.\log x$ $\left(\log \right(a^n)=n\log a)$

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

$\frac{d\left(\log w\right)}{dx}=\frac{d(x.\log x)}{dx}$

$\frac{d\left(\log w\right)}{dw}\left(\frac{dw}{dx}\right)=\frac{d(x.\log x)}{dx}$

$\frac{1}{w}\cdot \frac{dw}{dx}=\frac{d(x.\log x)}{dx}$

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

$\frac{1}{w}\cdot \left(\frac{dw}{dx}\right)=\frac{d\left(x\right)}{dx}.\log x+\frac{d\left(\log x\right)}{dx}\cdot x$

$\frac{1}{w}\cdot \left(\frac{dw}{dx}\right)=1.\log x+\frac{1}{x}.x$

$\frac{dw}{dx}=w(\log x+1)$

Substituing $w=x^x$, we get,

$\frac{dw}{dx}=x^x(\log x+1\cdots (iv\left)\right)$

Finding $\frac{dy}{dx}$

From (i)

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

Substituting the values of (ii), (iii) and (iv) in (i) then we get,

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

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

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