Course:MATH110/Archive/2010-2011/003/Notes/Computing derivatives using limits

Recall that the derivative, $f^{\prime }(a)$, of a function $f$ at a point $a$ is the slope of the tangent line to the the function at the point $x=a$. We'll be interested in this note on how to compute this using limits.

The function $f(x)=x^3-7x^2+14x-7$, its tangent line at the point ($1$,$1$) and a secant line through the points ($0.5$, $f(0.5)$) and ($1.5$,$f(1.5)$) to approximate the tangent line.

Consider for example the graph $y=f(x)$ of the function $f(x)=x^3-7x^2+14x-7$ and its tangent line at the point (1,1). What is the slope of the tangent line there.?

In this picture, it seems that the slope is 1, how can we make sure of this?

There are several ideas that might work. One that is popular, is to try to find some good approximations. How to do that? By constructing a line that is very similar to the tangent line (being similar isn't that great) but for which we can actually easily compute the slope (that's a good deal then).

More precisely, consider two points, one on the left, one on the right of $x=a$, both at distance $h$. This means the one on the left is now the point $x=a-h$ and the one on the right is the point $x=a+h$.

In our example, we chose to take the value $h=0.5$ and obtained the secant line as pictured.

Since we have these two points, we can compute the slope of the secant line to get an approximation of the slope of the tangent line. We obtain the following approximation:

$f^{\prime }(a)\approx \frac{f(a+h)-f(a-h)}{(a+h)-(a-h)}=\frac{f(a+h)-f(a-h)}{2h}$

Clearly, the small we let the value of $h$ be, the closer the secant line will be to the tangent line, the better the approximation. This is true and it can be proved that actually, if we let $h$ tend to zero, the limit of this approximation will precisely be the slope of the tangent line (which we can also call the derivative at this point). This can be written more concisely by:

$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a-h)}{2h}$

This is exactly what we wanted! The ability to compute the derivative of a function at a given point. We went from an interesting question: how to compute the slope of the tangent line to a function at a given point? to finding a solution to this problem by saying we can use limits to compute this precisely.

Example

Let's actually use what we just discussed to compute precisely the slope of the tangent line of the function $f(x)=x^2$ at the point $x=3$. Recall that the slope of the tangent line to the function $f$ at the point $x=3$ is denoted by $f^{\prime }(3)$ (it's just a notation, nothing else). So, we compute the following limit: $f^{\prime }(3)=\underset{h\to 0}{\lim }\frac{f(3+h)-f(3-h)}{2h}=\underset{h\to 0}{\lim }\frac{(3+h{)}^2-(3-h{)}^2}{2h}$ $=\underset{h\to 0}{\lim }\frac{(9+6h+h^2)-(9-6h+h^2)}{2h}=\underset{h\to 0}{\lim }\frac{12h}{2h}=6$ This means that the slope of the tangent line to the parabola $y=x^2$ at the point $x=3$ is 6.