Differentiation of the simple exponential function

Let $y = a^v$,         $a > 0$. Taking the logarithm of both sides to the base $e$, we get $\log y = v \log a$, or $v = \frac{\log y}{\log a} = \frac{1}{\log a} \cdot \log y$. Differentiate with respect to $y$ by formula (VIIIa),

$$\frac{dv}{dy} = \frac{1}{\log a} \cdot \frac{1}{y};$$

and from (5.2),relating to inverse functions, we get $\frac{dy}{dv} = \log a \cdot y$, or,

$$\frac{dy}{dv} = \log a \cdot a^v.$$

Since $v$ is a function of $x$ and it is required to differentiate $a^v$ with respect to $x$, we must use formula (5.1), for differentiating a function of a function, namely,

$$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}.$$

Substituting the value of $\frac{dy}{dx}$ from above, we get

$$\frac{dy}{dx} = \log a \cdot a^v \cdot \frac{dv}{dx}.$$

Therefore, $\frac{d}{dx} (a^v) = \log a \cdot a^v \cdot \frac{dv}{dx}$ (equation (IX) in §5.1 above). When $a = e$, $\log a = \log e = 1$, and (IX) becomes $\frac{d}{dx} (e^v) = e^v \frac{dv}{dx}$ (equation (IXa) in §5.1 above) .

The derivative of a constant with a variable exponent is equal to the product of the natural logarithm of the constant, the constant with the variable exponent, and the derivative of the exponent.