If x^y+y^x=b^a+a^b then find d y d x . - Vidyadip

Question

If $x^y+y^x=b^a+a^b$ then find $\frac{d y}{d x}$.

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Answer

ATQ $\quad x^y+y^x=b^a+a^b$
or $\quad u+v=b^a+a^b$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 w.r.t. $x$$
\begin{aligned}
\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 \quad \frac{d u}{d x} & =u\left[\left(\frac{d y}{d x}\right) \log x+\frac{y}{x}\right]
\end{aligned}
$

$\begin{array}{l}
\text { again } \quad \begin{array}{l}
v=y^x \\
\text { taking } \operatorname{logarithmic} \\
\log v=\log y^x=x \log y \quad\left[\because \log m^n=n \log m\right] \\
\log v=x \log y
\end{array}
\end{array}
$now differentiating w.r.t. $x$$
\begin{aligned}
& \\
\Rightarrow \quad & \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 \quad & \frac{d v}{d x}\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]
\end{aligned}$

now $\quad u+v=b^a+a^b$
$\therefore \quad \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 $\frac{d u}{d x}, \frac{d v}{d x}$
$x^y\left[(\log x) \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]=0$
$\begin{array}{l}\Rightarrow \quad\left(x^y \log x+\frac{x \cdot y^x}{y}\right) \frac{d y}{d x}+x^y \cdot \frac{y}{x}+y^x \log y=0 \\ \text { or } \quad\left(x^y \log x+x \cdot y^{x-1}\right) \frac{d y}{d x}+y \cdot x^{y-1}+y^x \log y=0 \\ \therefore \quad \quad \frac{d y}{d x}=-\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}\end{array}$

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