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

This is the general formula for differentiating from first principles.

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

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

Remember the expression in the brackets gives the gradient of the chord between \(x\) and \(x+h\). Taking the limit as \(h \to 0\) gives us the gradient of the function at \(x\).

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

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

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

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

The value of \(h\) is tending to \(0\), so any expressions not involving \(h\) are unaffected by this limiting process.

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

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

Think about how the expression \(\dfrac{a^h - a^0}{h}\) could be represented on the graph of \(f(x) = a^x\). What does it show?

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

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

The final step implies that \(\displaystyle{\lim_{h \to 0} \dfrac{a^h - a^0}{h} = f'(0)}\).

This is not a simple limit to find as both the numerator and denominator tend to zero as \(h \to 0\). However, if we look at how we find \(f'(x)\) from first principles we should see some similarities. \[f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \text{ and } \lim_{h \to 0} \dfrac{a^h - a^0}{h}\] Instead of finding the gradient for a generic point \(x\), the limit on the right gives us the gradient for the specific point \(x = 0\), i.e. \(f'(0)\).

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

- What does the graph of \(f'(x)\) look like?

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

The gradient function is the original function multiplied by a constant. This tells us the graph of \(f'(x)\) will look very similar to \(f(x).\)In fact it will be a transformation of the original function; a stretch in the \(y\) direction, scale factor \(f'(0).\) (Or a stretch in the \(x\) direction, scale factor \(\frac{1}{f'(0)}.\))

What do the graphs tell you about \(f'(0)\) when \(a = 2\) and when \(a = 4\)?

- How will the gradient function change as \(a\) changes?

We can see from the graphs above that as \(a\) gets bigger, \(f(x) = a^x\) gets steeper more quickly and so the gradient at zero will be larger. Therefore, as \(a\) gets bigger, \(f'(x)\) will also get steeper more quickly. What will happen as \(a\) gets smaller?

- There is a special case of the gradient function of \(a^x\). Can you think what it might be? To find out more look at A special case.

The gradient function itself could be steeper than the original function. This will happen if the stretch factor is bigger than one, i.e. \(f'(0) >1\). It could also be less steep, i.e. when \(f'(0) <1.\) This implies there will be a special case in the middle when \(f'(0) = 1\) and the gradient function and original function will be exactly the same!