Question

Find  \[\frac{dy}{dx}\] \[y = e^x + {10}^x + x^x\] ?

 

Answer 1

\[\text{We have, y} = e^x + {10}^x + x^x \]
\[ \Rightarrow y = e^x + {10}^x + e^{\log x^x} \]
\[ \Rightarrow y = e^x + {10}^x + e^{x\log x}\]

Differentiating with respect to x,

\[\frac{dy}{dx} = \frac{d}{dx}\left( e^x \right) + \frac{d}{dx}\left( {10}^x \right) + \frac{d}{dx}\left( e^{x\log x} \right)\]

\[ = e^x + {10}^x \log10 + e^{x\log x} \frac{d}{dx}\left( x \log x \right)\]
 \[ = e^x + {10}^x \log10 + e^{x\log x} \left[ x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right) \right] \]

\[ = e^x + {10}^x \log10 + e^{\log x^x }\left[ x\left( \frac{1}{x} \right) + \log x \right]\]

\[ = e^x + {10}^x \log10 + x^x \left[ 1 + \log x \right]\]

\[ = e^x + {10}^x \log10 + x^x \left[ \log e + \log x \right] ..........\left[ \because \log_e e = 1 \right]\]

\[ = e^x + {10}^x \log10 + x^x \left( \log e x \right) ............\left[ \because \log A + \log B = \log AB \right]\]