Question

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

 

Answer 1

\[\text{ We have, y } = e^{x^{e^x}} + x^{e^{e^x}} + e^{x^{x^e}} \]
\[ \Rightarrow y = u + v + w\]
\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} . . . \left( i \right)\]
\[\text{ where u } = e^{x^{e^x}} , v = x^{e^{e^x}} \text{ and w } = e^{x^{x^e}} \]
\[\text{ Now, u } = e^{x^{e^x}} . . . \left( ii \right)\]

Taking log on both sides,

\[\log u = \log e^{x^{e^x}} \]
\[ \Rightarrow \log u = x^{e^x} \log e\]
\[ \Rightarrow \log u = x^{e^x} . . . \left( iii \right)\]

Taking log on both sides,

\[\log \log u = \log x^{e^x} \]
\[ \Rightarrow \log \log u = e^x \log x\]

Differentiating with respect to x,

\[\Rightarrow \frac{1}{\log u}\frac{d}{dx}\left( \log u \right) = e^x \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{1}{\log u}\frac{1}{u}\frac{du}{dx} = \frac{e^x}{x} + e^x \log x\]
\[ \Rightarrow \frac{du}{dx} = u\log u\left[ \frac{e^x}{x} + e^x \log x \right]\]
\[ \Rightarrow \frac{du}{dx} = e^{x^{e^x}} \times x^{e^x} \left[ \frac{e^x}{x} + e^x \log x \right] . . . \left( A \right)\]
\[ \left[ \text{ Using equation } \left( ii \right) \text{ and } \left( iii \right) \right]\]
\[\text{ Now, v } = x^{e^{e^x}} . . . \left( iv \right)\]

Taking log on both sides,

\[\log v = \log x^{e^{e^x}} \]
\[ \Rightarrow \log v = e^{e^x} \log x\]

\[\Rightarrow \frac{1}{v}\frac{dv}{dx} = e^{e^x} \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^{e^x} \right)\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = e^{e^x} \left( \frac{1}{x} \right) + \log x e^{e^x} \frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ e^{e^x} \left( \frac{1}{x} \right) + \log x e^{e^x} e^x \right]\]
\[ \Rightarrow \frac{dv}{dx} = x^{e^{e^x}} \times e^{e^x} \left[ \frac{1}{x} + e^x \log x \right] . . . \left( B \right) \]
\[ \left\{ \text{ Using equation } \left( 4 \right) \right\}\]
\[\text{ Now, w } = e^{x^{x^e}} . . . \left( v \right)\]

Taking log on both sides,

\[\log w = \log e^{x^{x^e}} \]
\[ \Rightarrow \log w = x^{x^e} \log e\]
\[ \Rightarrow \log w = x^{x^e} . . . \left( vi \right)\]

Taking log on both sides,

\[\log \log w = \log x^{x^e} \]
\[ \Rightarrow \log \log w = x^e \log x\]

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