Question

Find $\frac{dy}{dx}$

$y=x^n+n^x+x^x+n^n$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=x^n+n^x+x^x+n^n \\ \Rightarrow y=x^n+n^x+e^{\log x^x}+n^n \\ \Rightarrow y=x^n+n^x+e^{x\log x}+n^n \end{gathered}$$
Differentiate with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(x^n\right)+\frac{d}{dx}\left(n^x\right)+\frac{d}{dx}\left(e^{x\log x}\right)+\frac{d}{dx}\left(n^n\right) \\ =n{x}^{n-1}+n^x\log n+e^{\log x^x}\left[x\frac{d}{dx}\log x+\log x\frac{d}{dx}\left(x\right)\right] \\ =n{x}^{n-1}+n^x\log n+x^x\left[x\left(\frac{1}{x}\right)+\log x\right] \\ =n{x}^{n-1}+n^x\log n+x^x\left[1+\log x\right] \\ =n{x}^{n-1}+n^x\log n+x^x\left[\log e+\log x\right]\left[\because {\log }_{e}e=1\mathrm{and}\log A+\log B=\log \left(AB\right)\right] \\ =n{x}^{n-1}+n^x\log n+x^x\log \left(ex\right) \end{gathered}$$