Question

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

Answer 1

\[\text{ We have, y }= x^n + n^x + x^x + n^n \]

\[ \Rightarrow y = x^n + n^x + e^{\log x^x }+ n^n \]

\[ \Rightarrow y = x^n + n^x + e^{x\log x} + n^n\]

Differentiate with respect to x,

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

\[ = n x^{n - 1} + n^x \log n + e^{\log x^x} \left[ x\frac{d}{dx}\log x + \log x\frac{d}{dx}\left( x \right) \right]\]

\[ = n x^{n - 1} + n^x \log n + x^x \left[ x\left( \frac{1}{x} \right) + \log x \right]\]

\[ = n x^{n - 1} + n^x \log n + x^x \left[ 1 + \log x \right]\]

\[ = n x^{n - 1} + n^x \log n + x^x \left[ \log e + \log x \right] .........\left[ \because \log_e e = 1 \text{ and }\log A + \log B = \log\left( AB \right) \right]\]

\[ = n x^{n - 1} + n^x \log n + x^x \log\left( ex \right)\]