Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 55

Answer:   $\frac{d y}{d x}=\frac{-\left(y^{x} \log y+x^{y-1} \cdot y+x^{x}(\log x+1)\right)}{\left(x^{y} \log y+y^{x-1} \cdot x\right)}$

Hint: To solve this equation we solve this differently

Given: $y^{x}+x^{y}+x^{x}=a^{b}$

Solution:  Let uvw

$u+v+w=a^{b}$

Diff w.r.t. x

\begin{aligned} &\frac{d}{d x}(u+v+w)=\frac{d}{d x} a^{b} \\\\ &\frac{d u}{d x}+\frac{d v}{d u}+\frac{d w}{d v}=0 \\\\ &u=y^{x} \end{aligned}            ...........(1)

Taking log on both sides,

$\log u=\log y^{x} \quad\left[\because \log a^{b}=b \log a\right]$

Diff w.r.t. u                              $\left[\because \frac{d}{d x}(u v)=u v^{\prime}+v u^{\prime}\right]$

$\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log y) \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}\right]$

\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y(1) \\\\ &\frac{d u}{d x}=u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \end{aligned}

$v=x^{y}$                                ..............(2)

Taking log on both sides,

\begin{aligned} &\log v=\log x^{y} \\\\ &\log v=y \log x \end{aligned}

Diff w.r.t

\begin{aligned} &\frac{1}{v} \frac{d v}{d x}=y \cdot \frac{1}{x}+\log x \frac{d y}{d x} \\\\ &\frac{d v}{d u}=v\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ &w=x^{x} \end{aligned}        ..........(3)

Log on b.s.

\begin{aligned} &\log w=\log x^{x} \\\\ &\log w=x \log x \end{aligned}

Diff w.r.t

$\frac{1}{w} \frac{d w}{d x}=x \cdot \frac{1}{x}+\log x(1)$

From (1)

$\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0$

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

$y^{x} \log y+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}+x^{y-1} \frac{y}{x}+x^{x}(1+\log x)=0$

$y^{x} \log y+x^{y} \cdot \frac{y}{x}+x^{x}(\log x+1)+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}=0$

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

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