# Is d/dx (a^x)=x a^{x-1}? (a conceptual error in O/A Level Math)

In O Level, students are taught that $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$

So naturally, students may think that $\displaystyle\frac{d}{dx}(a^{x})=xa^{x-1}??$ (a is a constant)

Well, actually that is good pattern spotting, but unfortunately it is incorrect. Do not be too disheartened if you make this mistake, it is a common mistake.

The above is a conceptual error as  $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$ only holds when n is a constant.

Fortunately, this question is rarely tested, though it is quite possible that it can come up in A Levels.

To fully understand the following steps, it would help read my other post (Why is e^(ln x)=x?) first.

First, we write $\displaystyle a^x=e^{\ln a^x}=e^{x\ln a}$.

Hence
$\displaystyle \begin{array}{rcl} \frac{d}{dx} (a^x)&=&\frac{d}{dx} (e^{x\ln a})\\ &=&e^{x\ln a}(\ln a)\\ &=&e^{\ln a^x}(\ln a)\\ &=&a^x(\ln a) \end{array}$

After fully understanding the above steps, you may memorize the formula if you wish:

$\boxed{\frac{d}{dx} (a^x)=a^x(\ln a)}$

Memory Tip: If you let a=e, you should get $\boxed{\frac{d}{dx} (e^x)=e^x(\ln e)=e^x}$

The above steps involve the chain rule, which I will cover in a subsequent post.

## Author: mathtuition88

https://mathtuition88.com/

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