Derivative of $a^{x^{1/2}}$

Question

I know that derivative of $a^x$ is $\ln(a)a^x$ but is there a rule which applies to $a^{x^{1/2}}$ or do I have to manipulate power of x with taking ln of both sides in order to find it?

Answer 1

This would be a simple Chain Rule: $$\begin{align*} \frac{d}{dx} a^{x^{1/2}} &= \ln(a) a^{x^{1/2}}(x^{1/2})’\\ &= \frac{1}{2}x^{-1/2}\ln(a)a^{x^{1/2}}. \end{align*}$$

Alternatively, recall that $a^b = e^{b\ln(a)}$. Thus, $$a^{x^{1/2}} = e^{(x^{1/2})\ln(a)}$$ so $$\begin{align*} \frac{d}{dx} a^{x^{1/2}} &= \frac{d}{dx} e^{x^{1/2}\ln(a)}\\ &= (x^{1/2}\ln(a))’e^{x^{1/2}\ln(a)}\\ &= \frac{1}{2}x^{-1/2}\ln(a)(a^{x^{1/2}}). \end{align*}$$

Alternatively, we can use logarithmic differentiation. Letting $y=a^{x^{1/2}}$, we have $$\begin{align*} \ln y &= x^{1/2}\ln(a)\\ \frac{d}{dx}(\ln y) &= \frac{d}{dx}(x^{1/2}\ln(a))\\ \frac{y’}{y} &= \frac{1}{2}x^{-1/2}\ln(a)\\ y’ &=\frac{1}{2}x^{-1/2}\ln(a)y\\ y’ &= \frac{1}{2}x^{-1/2}\ln(a)(a^{x^{1/2}}). \end{align*}$$

In general, by the Chain Rule, $$\frac{d}{dx} a^{f(x)} = \ln(a)a^{f(x)}f’(x).$$

Answer 2

Using the "chain rule", as Eevee Trainer suggested, let $y= x^{1/2}$. The $f(y)= a^y$ so $\frac{df}{dy}= a^y \ln(a)= a^{x^{1/2}} \ln(a)$ and $\frac{dy}{dx}= \frac{1}{2}x^{-1/2}$. The chain rule then says that $\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}= \left(a^{x^{1/2}} (a)\right)\left(\frac{1}{2}x^{-1/2}\right)= \frac{1}{2} \ln(a) x^{-1/2}a^{x^{1/2}}$