#### Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 7 maths textbook solution

\begin{aligned} &\text { Answer: } \frac{d y}{d x}=e^{x^{e^{x}}} \cdot x^{e^{x}}\left\{\frac{e^{x}}{x}+e^{x} \log \log x\right\}+x^{e^{e^{x}}} \cdot e^{e^{x}}\left\{\frac{1}{x}+e^{x} \log \log x\right\}+ \\ &e^{x^{x^{e}}} \cdot x^{x^{e}} \cdot x^{e-1}\{1+e \quad \log \log x\} \end{aligned}

Hint: Let u =$e^{x^{e^{x}}}$, v =$x^{x^{e^{x}}}$ and w = $e^{x^{e^{x}}}$

Find$\inline \frac{du}{dx}$, $\inline \frac{dv}{dx}$ & $\inline \frac{dw}{dx}$ to get $\inline \frac{dy}{dx}$

Given: $\inline y=e^{x^{e^{x}}}+x^{x^{e^{x}}}+e^{x^{e^{x}}}$

Solution:

Here it is given that,

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

Let us consider u =$\inline e^{x^{e^{x}}}$, v = $\inline x^{x^{e^{x}}}$and w = $\inline e^{x^{e^{x}}}$

So, y = u + v + w & $\inline \frac{dy}{dx}$= $\inline \frac{du}{dx}$+$\inline \frac{dv}{dx}$+$\inline \frac{dw}{dx}$

Now, u = $\inline e^{x^{e^{x}}}$

Taking log on both sides, we get:

\inline \begin{aligned} \log u &=\log e^{x^{x^{x}}} \\ &=x^{e^{x}} \log e \\ \end{aligned}

\inline \begin{aligned} \log u &=x^{e^{x}} \\ \end{aligned}                                                    \inline \begin{aligned} (\therefore \log e=1) \end{aligned}                              …(1)

Again taking log on both sides,

\inline \begin{aligned} &\log (\log u)=\log x^{e^{x}} \\ &\therefore \log (\log u)=e^{x} \log x \end{aligned}

…(2)

Differentiate (2) w.r.t x,

\inline \begin{aligned} &\frac{1}{\log u} \cdot \frac{1}{u} \frac{d u}{d x}=e^{x} \cdot \frac{1}{x}+e^{x} \cdot \log x \\ &\frac{d u}{d x}=\left(e^{x} \cdot \frac{1}{x}+e^{x} \cdot \log x\right) u \cdot \log u \\ &\frac{d u}{d x}=\left(e^{x} \cdot \frac{1}{x}+e^{x} \cdot \log x\right) \cdot e^{x^{x^{x}}} \cdot x^{e^{x}} \end{aligned}                                                                ...(A)

Now, v = $e^{x^{e^{x}}}$

Taking log on both sides, we get:

\inline \begin{aligned} &\log v=\log x^{e^{e^{x}}} \\ &\log v=e^{e^{x}} \log x \end{aligned}                                                                                                    …(3)

Differentiate (3) w.r.t x,

\inline \begin{aligned} &\frac{1}{v} \frac{d v}{d x}=e^{e^{x}} \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(e^{e^{x}}\right) \\ &\frac{1}{v} \frac{d v}{d x}=e^{e^{x}} \frac{1}{x}+\log x \cdot\left(e^{e^{x}}\right) \cdot e^{x} \end{aligned}

\inline \begin{aligned} &\frac{d v}{d x}=\left[e^{e^{x}} \frac{1}{x}+\log x \cdot\left(e^{e^{x}}\right) \cdot e^{x}\right] v \\ &\frac{d v}{d x}=e^{e^{x}}\left[\frac{1}{x}+\log x \cdot\left(e^{x}\right)\right] x^{e^{e^{x}}} \end{aligned}                                                                             …(B)

Now, w = $e^{x^{e^{x}}}$

Taking log on both sides, we get:

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

$\inline \log w=x^{x^{x}}$                                                                    $\inline \quad(\therefore \log e=1)$                           …(4)

Again taking log on both sides,

\inline \begin{aligned} &\log (\log w)=\log x^{x^{2}} \\ &\therefore \log (\log w)=x^{e} \log x \end{aligned}                                                                                                  …(5)

Differentiate (5) w.r.t x,

\inline \begin{aligned} &\frac{1}{\log w} \cdot \frac{d}{d x}(\log w)=x^{e} \cdot \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(x^{e}\right) \\ &\frac{1}{\log w} \cdot \frac{1}{w} \frac{d w}{d x}=x^{e} \cdot \frac{1}{x}+\log x\left(e x^{e-1}\right) \\ &\frac{d w}{d x}=w \log w\left[x^{e-1}+\log x\left(e x^{e-1}\right)\right] \\ &\frac{d w}{d x}=e^{x^{x^{*}}} \cdot x^{x^{\prime}} \cdot x^{e-1}[1+e \log (x)] \end{aligned}                                                                     …(C)

Using A,B and C, $\inline \frac{d y}{d x}$becomes:

\inline \begin{aligned} \frac{d y}{d x}=e^{x^{e^{x}}} \cdot x^{e^{x}} &\left\{\frac{e^{x}}{x}+e^{x} \log \log x\right\}+x^{e^{e^{x}}} \cdot e^{e^{x}}\left\{\frac{1}{x}+e^{x} \log \log x\right\} \\ &+e^{x^{x^{e}}} \cdot x^{x^{e}} \cdot x^{e-1}\{1+e \quad \log \log x \end{aligned}

Hence, it is proved.