Question

Find $\frac{dy}{dx}$

$y=e^x+{10}^x+x^x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=e^x+{10}^x+x^x \\ \Rightarrow y=e^x+{10}^x+e^{\log x^x} \\ \Rightarrow y=e^x+{10}^x+e^{x\log x} \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^x\right)+\frac{d}{dx}\left({10}^x\right)+\frac{d}{dx}\left(e^{x\log x}\right) \\ =e^x+{10}^x\log 10+e^{x\log x}\frac{d}{dx}\left(x\log x\right) \\ =e^x+{10}^x\log 10+e^{x\log x}\left[x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)\right] \\ =e^x+{10}^x\log 10+{e^{\log x}}^x\left[x\left(\frac{1}{x}\right)+\log x\right] \\ =e^x+{10}^x\log 10+x^x\left[1+\log x\right] \\ =e^x+{10}^x\log 10+x^x\left[\log e+\log x\right]\left[\because {\log }_{e}e=1\right] \\ =e^x+{10}^x\log 10+x^x\left(\log ex\right)\left[\because \log A+\log B=\log AB\right] \end{gathered}$$