Question

If $y=e^{x^{e^x}}+x^{e^{e^x}}+e^{x^{x^e}}$, prove that $\frac{dy}{dx}=e^{x^{e^x}}·x^{e^x}\left\{\frac{e^x}{x}+e^x·\log x\right\}$
$+x^{e^{e^x}}·e^{e^x}\left\{\frac{1}{x}+e^x·\log x\right\}+e^{x^{x^e}}{x}^{x^e}·x^{e-1}\left\{x+e\log x\right\}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=e^{x^{e^x}}+x^{e^{e^x}}+e^{x^{x^e}} \\ \Rightarrow y=u+v+w \\ \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}...\left(i\right) \\ \mathrm{where}u=e^{x^{e^x}},v=x^{e^{e^x}}\mathrm{and}w=e^{x^{x^e}} \\ \mathrm{Now},u=e^{x^{e^x}}...\left(ii\right) \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log u=\log e^{x^{e^x}} \\ \Rightarrow \log u=x^{e^x}\log e \\ \Rightarrow \log u=x^{e^x}...\left(iii\right) \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log \log u=\log x^{e^x} \\ \Rightarrow \log \log u=e^x\log x \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \Rightarrow \frac{1}{\log u}\frac{d}{dx}\left(\log u\right)=e^x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(e^x\right) \\ \Rightarrow \frac{1}{\log u}\frac{1}{u}\frac{du}{dx}=\frac{e^x}{x}+e^x\log x \\ \Rightarrow \frac{du}{dx}=u\log u\left[\frac{e^x}{x}+e^x\log x\right] \\ \Rightarrow \frac{du}{dx}=e^{x^{e^x}}\times x^{e^x}\left[\frac{e^x}{x}+e^x\log x\right]...\left(A\right) \\ \left[\mathrm{Using}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right)\right] \\ \mathrm{Now},v=x^{e^{e^x}}...\left(iv\right) \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log v=\log x^{e^{e^x}} \\ \Rightarrow \log v=e^{e^x}\log x \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{v}\frac{dv}{dx}=e^{e^x}\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(e^{e^x}\right) \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=e^{e^x}\left(\frac{1}{x}\right)+\log x{e}^{e^x}\frac{d}{dx}\left(e^x\right) \\ \Rightarrow \frac{dv}{dx}=v\left[e^{e^x}\left(\frac{1}{x}\right)+\log x{e}^{e^x}{e}^x\right] \\ \Rightarrow \frac{dv}{dx}=x^{e^{e^x}}\times e^{e^x}\left[\frac{1}{x}+e^x\log x\right]...\left(B\right) \\ \left\{\mathrm{Using}\mathrm{equation}\left(4\right)\right\} \\ \mathrm{Now},w=e^{x^{x^e}}...\left(v\right) \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log w=\log e^{x^{x^e}} \\ \Rightarrow \log w=x^{x^e}\log e \\ \Rightarrow \log w=x^{x^e}...\left(vi\right) \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log \log w=\log x^{x^e} \\ \Rightarrow \log \log w=x^e\log x \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{\log w}\frac{d}{dx}\left(\log w\right)=x^e\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x^e\right) \\ \Rightarrow \frac{1}{\log w}\left(\frac{1}{w}\right)\frac{dw}{dx}=x^e\left(\frac{1}{x}\right)+\log xe{x}^{e-1} \\ \Rightarrow \frac{dw}{dx}=w\log w\left[x^{e-1}+e\log x{x}^{e-1}\right] \\ \Rightarrow \frac{dw}{dx}=e^{x^{x^e}}{x}^{x^e}{x}^{e-1}\left(1+e\log x\right)----\left(C\right) \\ \left[\mathrm{using}\mathrm{equation}\left(\mathrm{v}\right),\left(\mathrm{vi}\right)\right] \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{A}\right),\left(\mathrm{B}\right)\mathrm{and}\left(\mathrm{C}\right)\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=e^{x^{e^x}}{x}^{e^x}\left[\frac{e^x}{x}+e^x\log x\right]+x^{e^{e^x}}\times e^{e^x}\left[\frac{1}{x}+e^x\log x\right]+e^{x^{x^e}}{x}^{x^e}{x}^{e-1}\left(1+e\log x\right) \end{gathered}$$