A special case

We have found that the gradient function of \(a^x\) can be written as \[f'(x) = a^x\times f'(0),\]

i.e. the gradient function is the function itself multiplied by a constant, and this constant is the gradient of \(a^x\) at \(x = 0\). There will be a special case when \(f'(0) = 1\) as then the gradient function of \(a^x\) would be itself, \(a^x\).

Before using the applet, look at the values given. Approximately what value do you think \(a\) will take to give \(f'(0) \approx 1?\)

Use the slider to find the value of \(a\) when \(f'(0) \approx 1\).

Does the gradient function look as you expected?

A very steep curve where the negative side the y-values are very close to 0, the graph intercepts the y-axis at 1 and the gradient on the positive side is very steep and doesn't reach far past x-values of 3

We find that \(f'(0) \approx 1\) when \(a = 2.72\). This means that the gradient function becomes \(f'(x) = a^x\), so \(a^x\) and its gradient function are identical.

Of course this value of \(a\) is an approximation, but does it look familiar?

It turns out that the exact value of \(a\) for which the gradient function is the same as the original function is \(a = 2.7181...\), in fact it is \(a = e\). You have probably met \(e\) before as the exponential function \(e^x\) and as the base of natural logarithms. Here we have found that if \(f(x) = e^x\), then \(f'(x) = e^x\).

What about other values of \(a\)? To think about what \(f'(0)\) is for any value, you can look at Differentiating exponentials.