I am trying to understand the derivative of the function $G(d)=a^d$. I am told it is the following (lecturers comments on my work)

$$ \frac{\partial}{\partial d}(a^d) = \frac{\partial}{\partial d}(e^{d\log a}) = (\log a)a^d $$

I'm not sure where the exponential comes from. Looking at this site the rule is

$$ \frac{\partial}{\partial d}(a^d) = a^d \ln a $$

and converting from $\ln$ to $\log$ does not seem to help. Can someone point me in the right direction?

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    $\begingroup$ $\log$ and $\ln$ both denote logarithm to the base $e$. $\endgroup$ Apr 22, 2017 at 14:34
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    $\begingroup$ And $\log$ is universally used to denote $\log_e$ in mathematics, while $\ln$ is more common where people believe their calculators. $\endgroup$
    – Chappers
    Apr 22, 2017 at 14:39
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    $\begingroup$ I dont think that is a universal convention. In many U.S. high school and college textbooks, $\log$ is $\log_{10}$ while $\ln$ corresponds to $\log_e$ $\endgroup$ Apr 22, 2017 at 14:51
  • $\begingroup$ @Just_to_Answer Sure, but Chappers mentioned "in mathematics"... $\endgroup$
    – Did
    Apr 22, 2017 at 21:42
  • $\begingroup$ @Did Haha...fair enough. Since this discussion has been done quite a bit in math.stackexchange.com/questions/90594/…, I dont think it is worth rehashing. $\endgroup$ Apr 22, 2017 at 22:42

1 Answer 1


The issue is whether the base is the variable or the exponent is the variable. If the base was the variable, i.e., $d^a$ then the derivative would be $a \, d^{a-1}$, applying power rule.

If the variable $d$ is in the exponent, you are dealing with an exponential function. Derivative of $e^x$ is $e^x$, then combining with chain rule, you get the result your lecturer gives.

For details, we want to have an equivalent expression for $a^d$ with $e$ as the base (so that we can use the derivative of $e^x$). If we temporarily let $y=a^d$, taking natural log we get $\ln y = d \, \ln a$, and exponentiating back you get $y = e^{d \, \ln a}$. Now taking the derivative with respect to $d$ (applying derivative of $e^x$ along with chain rule) you get $\frac{dy}{dd} = e^{d \, \ln a} \, \ln a = a^d \, \ln a$



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