# Solve this

Question:

If $y=e^{x^{x^{x}}}+x^{e^{x}}+e^{x^{x^{e}}}$, prove that

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}^{\mathrm{e}^{\mathrm{x}}}} \cdot \mathrm{x}^{\mathrm{e}^{\mathrm{x}}}\left\{\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{e}^{\mathrm{x}} \cdot \log \mathrm{x}\right\}+\mathrm{x}^{\mathrm{e}^{\mathrm{x}}} \cdot \mathrm{e}^{\mathrm{e}^{\mathrm{x}}}$

$\left\{\frac{1}{x}+e^{x} \cdot \log x\right\}+e^{x^{x^{e}}} x^{x^{e}} \cdot x^{e-1}\{1+e \log x\}$

Solution:

Here,

$y=e^{e^{e^{x}}}+x^{e^{e^{x}}}+e^{e^{x^{e}}}$

$\mathrm{y}=\mathrm{U}+\mathrm{V}+\mathrm{W}$

$\frac{d y}{d x}=\frac{d U}{d x}+\frac{d V}{d x}+\frac{d Z}{d x}$        .....(1)

Where, $u=e^{x^{e^{x}}}, v=x^{e^{e^{x}}}, w=e^{x^{x^{e}}}$

$u=e^{e^{e^{x}}}$

Taking log on both sides,

$\log u=\log e^{x^{e^{x}}}$

$\log u=x^{e^{x}} \log e$

$\log u=x^{e^{x}}$

Again, Taking log on both sides,

$\log \log u=\log x^{e^{x}}$

$\log \log u=e^{x} \log x$

Differentiating both sides with respect to $x$ by using the product rule,

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

$\frac{1}{u} \frac{1}{\log u} \frac{d u}{d x}=e^{x} \frac{1}{x}+e^{x} \log x$

$\frac{d u}{d x}=u * \log u\left(\frac{e^{x}}{x}+e^{x} \log x\right)$

Put value of $u$ and $\log u$,

$\frac{d u}{d x}=e^{e^{e^{x}}} * x^{e^{x}}\left(\frac{e^{x}}{x}+e^{x} \log x\right) \ldots \ldots(A)$

Now,

$v=x^{e^{e^{x}}}$

taking log on both sides,

$\log v=\log _{x} e^{e^{x}}$

$\log v=e^{e^{x}} \log x$

Differentiating both sides with respect to $\mathrm{x}$ by using the product rule,

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

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

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

Put value of $\mathrm{v}$,

$\frac{d v}{d x}=x^{e^{e^{x}}}\left[e^{e^{x} \frac{1}{x}}+e^{x} \log x e^{e^{x}}\right] \ldots \ldots$(B)

Now,

$w=e^{x^{x^{e}}}$

taking log on both sides,

$\log w=\log _{e^{x^{x^{6}}}}$

$\log w=x^{x^{e}} \log e$

$\log w=x^{x^{6}}$

taking log both sides,

$\log \log w=x^{e} \log x$

Differentiating both sides with respect to $x$ by using the product rule,

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

$\frac{1}{w} \frac{1}{\log w} \frac{d w}{d x}=x^{e} \frac{1}{x}+x^{e-1} \log e$

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

Put the value of $w$ and $\log w$,

$\frac{d w}{d x}=e^{x^{e^{e}}} * x^{x^{e}}\left(x^{e-1}+e \log x x^{e-1}\right)$          .......(c)

Using equation $A, B$ and $C$ in equation (1),

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

$* x^{x^{e}}\left(x^{e-1}+e \log x x^{e-1}\right)$

Hence, proved.