Question

e^{x log a} + e^{a long x} + e^{a log a}

Answer 1

$$\begin{gathered} \frac{d}{dx}\left(e^{x\log a}+e^{a\log x}+e^{a\log a}\right) \\ =\frac{d}{dx}\left(e^{x\log a}\right)+\frac{d}{dx}\left(e^{a\log x}\right)+\frac{d}{dx}\left(e^{a\log a}\right) \\ =\frac{d}{dx}\left(e^{\log a^x}\right)+\frac{d}{dx}\left({e^{\log x}}^a\right)+\frac{d}{dx}\left(e^{\log a^a}\right) \\ =\frac{d}{dx}\left(a^x\right)+\frac{d}{dx}\left(x^a\right)+\frac{d}{dx}\left(a^a\right) \\ =a^x\log a+a{x}^{a-1}+0 \\ =a^x\log a+a{x}^{a-1} \end{gathered}$$