Question
If $x^y+y^x=b^a+a^b$ then find $\frac{d y}{d x}$.
✓
Answer
$\text{ATQ}$
$x^y+y^x=b^a+a^b$
or $ u+v=b^a+a^b$ when suppose $u=x^y, v=y^x$ taking logarithmic
$\log u=\log x^y=y \log x \quad\left[\log m^n=n \log m\right]$
now differentiating $\text{w.r.t. x}$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \frac{d}{d x}(\log x)$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \cdot \frac{1}{x}$
$\therefore \frac{d u}{d x} =u\left[\left(\frac{d y}{d x}\right) \log x+\frac{y}{x}\right]$
again
$v=y^x$
taking $\operatorname{logarithmic}$
$\log v=\log y^x=x \log y\ \left[\because \log m^n=n \log m\right]$
$\log v=x \log y$
now differentiating $\text{w.r.t. x}$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=1 \cdot \log y+x \cdot \frac{d}{d x}(\log y)$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log y+x \cdot \frac{1}{y} \frac{d y}{d x}$
$\Rightarrow \frac{d v}{d x}=\log y+\frac{x}{y} \frac{d y}{d x}$
$\therefore \frac{d v}{d x}\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]$
now $ u+v=b^a+a^b$
$\therefore \frac{d u}{d x}+\frac{d v}{d x}=\frac{d}{d x}\left(b^a+a^b\right)=0$
from equation $(1)$ and $(2),$
Put the value of $\frac{d u}{d x}, \frac{d v}{d x}$
$x^y\left[(\log x) \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]=0$
$\Rightarrow \quad\left(x^y \log x+\frac{x \cdot y^x}{y}\right) \frac{d y}{d x}+x^y \cdot \frac{y}{x}+y^x \log y=0$
$\text { or } \quad\left(x^y \log x+x \cdot y^{x-1}\right) \frac{d y}{d x}+y \cdot x^{y-1}+y^x \log y=0$
$\therefore \frac{d y}{d x}=-\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}$
$x^y+y^x=b^a+a^b$
or $ u+v=b^a+a^b$ when suppose $u=x^y, v=y^x$ taking logarithmic
$\log u=\log x^y=y \log x \quad\left[\log m^n=n \log m\right]$
now differentiating $\text{w.r.t. x}$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \frac{d}{d x}(\log x)$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \cdot \frac{1}{x}$
$\therefore \frac{d u}{d x} =u\left[\left(\frac{d y}{d x}\right) \log x+\frac{y}{x}\right]$
again
$v=y^x$
taking $\operatorname{logarithmic}$
$\log v=\log y^x=x \log y\ \left[\because \log m^n=n \log m\right]$
$\log v=x \log y$
now differentiating $\text{w.r.t. x}$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=1 \cdot \log y+x \cdot \frac{d}{d x}(\log y)$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log y+x \cdot \frac{1}{y} \frac{d y}{d x}$
$\Rightarrow \frac{d v}{d x}=\log y+\frac{x}{y} \frac{d y}{d x}$
$\therefore \frac{d v}{d x}\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]$
now $ u+v=b^a+a^b$
$\therefore \frac{d u}{d x}+\frac{d v}{d x}=\frac{d}{d x}\left(b^a+a^b\right)=0$
from equation $(1)$ and $(2),$
Put the value of $\frac{d u}{d x}, \frac{d v}{d x}$
$x^y\left[(\log x) \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]=0$
$\Rightarrow \quad\left(x^y \log x+\frac{x \cdot y^x}{y}\right) \frac{d y}{d x}+x^y \cdot \frac{y}{x}+y^x \log y=0$
$\text { or } \quad\left(x^y \log x+x \cdot y^{x-1}\right) \frac{d y}{d x}+y \cdot x^{y-1}+y^x \log y=0$
$\therefore \frac{d y}{d x}=-\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}$
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