To the limit

To the limit

Excerpt: Using a series of statements, students are asked to put together a chain of reasoning to find the derivative of $a^x$. This leads on to looking at a special case of this derivative.

Warm-up

What is the gradient of the chord $AB$?

Solution

$$\begin{align*} \text{gradient of } AB &= \dfrac{f(x+h) - f(x)}{x+h - x} \\ &=\dfrac{a^{x+h} - a^x}{h} \end{align*}$$

This expression gives us the gradient of $AB.$ If $h$ is small enough, then this also gives us an estimation of the gradient of $f(x)$ at $A.$

How could we find the exact gradient at $A$? See the gradient of $a^x$ to find out.

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.

Solution

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

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!

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 $f'(0) \approx 1?$

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

Does the gradient function look as you expected?

Solution

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.