Checking derivation of y = a^x

Question

Can you tell me if there are any flaws with this derivation of $y = a^x$...

The assumptions are that the derivative $$\frac{d}{dx}e^x = e^x$$ and that the derivative $$\frac{d}{dx}\ln x = \frac{1}{x} = x^-1.$$

$$\begin{array}{rcl} y &=& a^x\\ \ln y &=& \ln a^x\\ \ln y &=& x \ln a\\ x &=& \frac 1{\ln (a) }\ln y\\ \frac {dx}{dy} &=& \frac {1}{\ln (a) } y^-1\\ \frac {dy}{dx} &=& \ln(a)y\\ \frac {d}{dx} a^x &=& \ln(a)a^x \end{array} $$ I find the natural logarithm in this result very odd, a bit random.

Answer 1

Your work is correct. Notice also that we can differentiate this function by simpler way:

$$y=a^x=e^{\ln(a) x}$$ so using the chain rule $$\frac{dy}{dx}=\ln(a)e^{\ln(a) x}=\ln(a) a^x$$

Answer 2

We start with y=a^x. Taking natural logarithms of both sides. ln⁡y=ln⁡〖a^x 〗 using the “3rd” logarithm law. ln⁡y=x ln⁡a Writing both sides with the base of e. y=e^(x ln⁡a ). Note this also equals a^x. Now all we do is differentiate using d/dx (e^f(x) )=f^' (x)e^(f(x)). y^'=ln⁡〖ae^(x ln⁡a ) 〗 or y^'=ln⁡〖a(a^x ).〗 ∎