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Question:

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

Solution:

Given that, $y^{x}+x^{y}+x^{x}=a^{b}$

Putting, $u=y^{x}, v=x^{y}, w=x^{x}$, we get

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

Therefore, $\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0$ .......(i)

Now, $u=y^{x}$,

Taking log on both sides, we have

$\log u=x \log y$b

Differentiating both sides with respect to $x$, we have

$=x \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot 1$

So, $\frac{d u}{d x}=u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)$

$=y^{x}\left[\frac{x}{y} \frac{d y}{d x}+\log y\right] \ldots \ldots$(2)

Also, $\mathrm{V}=\mathrm{X}^{y}$,

Taking log on both sides, we have

$\log v=y \log x$

Differentiating both sides with respect to $x$, we have

$\frac{1}{v} \frac{d v}{d x}=y \frac{d}{d x}(\log x)+\log x \frac{d y}{d x}$

$=y \frac{1}{x}+\log x \frac{d y}{d x}$

So, $\frac{d v}{d x}=v\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)$

$=x^{y}\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \ldots \ldots$ (iii)

Again, $w=x^{x}$,

Taking log on both sides, we have

$\log w=x \log x$

Differentiating both sides with respect to $x$, we have

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

$=x \cdot \frac{1}{x}+\log x \cdot 1$

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

$=x^{x}[1+\log x] \ldots \ldots$ (iv)

From (i), (ii), (iii), (iv)

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

$y^{x}\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]+x^{y}\left[\frac{y}{x}+\log x \frac{d y}{d x}\right]+x^{x}[1+\log x]=0$

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

Therefore,

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

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