We already know that the exponential function \(e^x\) is its own tangent/gradient function. However, there are other exponential functions such as \(2^x\), \(3^x\) and so on.
Are their tangent/gradient functions the same or does something else happen?
What can we say about \(1^x\) or \(0.5^x\), for example?
Here is a series of structured questions leading to finding the derivative of \(a^x\) with respect to \(x\) for a positive number \(a\).
If \(e^k=2\), what is \(k\)?
We can apply the natural logarithm to both sides of our equation to find that \(\ln e^k =\ln 2\). Using the laws of logarithms we find \[\ln e^k = k\ln e = k,\] so that \(k=\ln 2\).
Which is steeper, \(e^x\) or \(2^x\)?
Can we write \(2^x\) in the form \(e^{kx}\), where \(k\) is a real number?
We can write \(2\) as \(e^{\ln 2}\) and so \(2^x=(e^{\ln 2})^x=e^{x\ln 2}\).
Does this support the idea that \(e^x\) is steeper than \(2^x\)?
What is the derivative of \(e^{kx}\) with respect to \(x\), where \(k\) is a real number?
Using the substitution \(u=kx\) and the knowledge that \(e^u\) when differentiated with respect to \(u\) is \(e^u\), the chain rule gives
\[\frac{d}{dx}(e^{kx})=\frac{d}{du}(e^{u}) \times \frac{du}{dx}=e^u \times k = ke^{kx}.\]
How can we differentiate \(2^x\) with respect to \(x\)?
We can rewrite \(2^x\) as \(e^{x\ln 2}\) and differentiate \(e^{x\ln 2}\) using our answer to the question above (since \(\ln 2\) is a real number). So we have the following:
