We can find the gradient of a function by taking the limit of the gradient of a chord between \(x\) and \(x+h\) as \(h\) tends to \(0\).

\[f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}\]

These statements and explanations can be used to create a chain of reasoning to find the gradient function of \(f(x) = a^x\). Sort the statements into a logical argument, either writing your own explanations of the steps, or turning over the hidden cards to use the explanations provided.

\(\displaystyle{f'(x) = \lim_{h \to 0} \left(a^x \left(\dfrac{a^{h} - 1}{h} \right) \right)}\)

The expression is factorised as \(a^x\) is a common factor.

\(\displaystyle{f'(x) = a^x \, \lim_{h \to 0} \left(\dfrac{a^{h} - a^0}{h} \right)}\)

As \(h \to 0\) the limit will give the gradient function at \(x = 0\).

\(\displaystyle{f'(x) = \lim_{h \to 0} \left(\dfrac{a^{x+h} - a^x}{h} \right)}\)

The expression in the limit gives the gradient of a chord between \(x = 0\) and \(x = h\) on \(f(x) = a^x\).

\(\displaystyle{f'(x) = a^x \, \lim_{h \to 0} \left(\dfrac{a^{h} - 1}{h} \right)}\)

This is the general formula for differentiating from first principles.

\(f'(x) = a^x \times f'(0)\)

This is an expression for the gradient of \(f(x) = a^x.\)

\(\displaystyle{f'(x) = \lim_{h \to 0} \left(\dfrac{f(x+h)-f(x)}{h} \right)}\)

As \(a^x\) is not affected by \(h\) it can be taken outside of the limit.

What does the final statement tell you about the gradient function of \(f(x)\)?

- What does the graph of \(f'(x)\) look like?
- How will the gradient function change as \(a\) changes?
- There is a special case of the gradient function of \(a^x\). What could it be? To find out more look at A special case.