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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