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The h-Form of the Derivative and Finding Tangent Lines

Business Calculus The Derivative as a Limit of an Average Rate of Change The h-Form of the Derivative and Finding Tangent Lines

A new way of computing derivatives

Recall the formula we used for computing the derivative of a function at a given $x$-value:

$$f'(c)=\lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c}$$

This formula is rather limited in terms of the functions it helps us compute derivatives for. For instance, suppose we want to compute the derivative of $f(x)=x^3$ at $x=2$. The formula above would give us (after a little simplifying):

$$f'(1)=\lim_{x\rightarrow 2} \frac{x^3-8}{x-2}$$

It isn’t at all clear how we can clean up this fraction so that we don’t get division by zero when we plug in $x=2$ to compute the limit. Perhaps we can rearrange the original formula so we have something more practical that enables us to carry out computations more easily.

In the formula $f'(c)=\lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c}$, notice that as $x$ approaches $c$, the “distance” (or rather the difference) between $x$ and $c$ becomes infinitely small; i.e. $x-c $ approaches $0$. Let’s let $h=x-c$. Then as $x\rightarrow c$, $h\rightarrow 0$. Solving for $x$ in the little equation $h=x-c$ gives us $x=h+c$.

So let’s rewrite the original derivative limit formula in terms of $h$ and $c$ so there are no more $x$’s anywhere. This gives us the following.

$h$-Form of the Derivative at a Given $x$-Value

Let $f$ be a function defined on an interval $I$ and let $x=c$ be an $x$-value inside this interval. The (first) derivative of $f$ at $x=c$, denoted $f'(c)$, is given by the following:
$$f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c)}{h}$$

Again, all we really did to get this new formula is replace $x-c$ with $h$ in the denominator, replace $x$ with $c+h$ inside $f$ in the numerator, and change up the limit from $\lim_{x\rightarrow c}$ to $lim_{h\rightarrow 0}$ because $h$ goes to $0$ when $x$ goes to $c$.

This new form helps us compute derivatives much more easily. The steps are as follows

How to compute derivatives using the $h$ form.

To find the derivative of a function $f$ at a given $x$-value $c$:

1.) Compute $f(c+h)$ where $c$ is the given x-value.

2.) Compute and simplify $f(c+h)-f(c)$

3.) Compute and simplify $\frac{f(c+h)-f(c)}{h}$

4.) Take the limit $\lim_{h\rightarrow 0}$ of the answer to the above.

Using the new formula:

$$f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c)}{h}$$

First, compute $f(c+h)$ where $c=2$. This gives us

$$\begin{align}f(2+h)&=(2+h)^3\\&=h^3+6h^2+12h+8\end{align}$$

Now we compute the entire numerator of that formula above: $f(c+h)-f(c)$. Note that, using what we just got for $f(2+h)$,

$$\begin{align}f(2+h)-f(2)&=(h^3+6h^2+12h+8)-(8)\\&=h^3+6h^2+12h\end{align}$$

Now, we compute $\frac{f(c+h)-f(c)}{h}$ by using what we last computed as our numerator. After a bit of factoring in the numerator, we can kill the $h$ in the denominator. Hence,

$$\begin{\frac{f(2+h)-f(2)}{h}&=\frac{h^3+6h^2+12h}{h}\\&=\frac{h\cdot(h^2+6h+12)}{h}\\&=h^2+6h+12\end{align}$$

Lastly, we take the limit of this result to get the full derivative computation:

$$\begin{align}f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c)}{h}&=\lim_{h\rightarrow 0} h^2+6h+12\\&=(0)^2+6(0)+12\\&=12\end{align}$$

Finding Tangent Lines Using the New Formula…

… works exactly the same way as before. We just use the new $h$-form of the derivative to compute the slope of the tangent line. Nothing else changes.

Compute $f(c+h)$ where $c$ is the given x-value.