## Gradient of $a^x$

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.