Derivative

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Julia: Ah, this sucks!
Dylan: What’s up?
Julia: I’m supposed to find the slope of a parabola at a point, and I’m not sure how!
Dylan: Well if we had two points we could make a secant line to approximate it!
Julia: Secant line? What’s that?
Dylan: A secant line is just a line which connects two points on a function!
Julia: But isn’t the tangent line one that skims a curve at one point? So the slope of the tangent line is the slope at that point! See? $\graph {f(x)=x^2,g(x)=2(x-1)+1}$
Dylan: Well do you know how to find the equation for a line with just one point?
Julia: ...
James: Come on guys we can approximate the tangent line using the secant line!
Altogether: Let’s dive in!

Guided Example

Consider the function $f(x) = x^2$ . In green is $g(x)=4x-4$ the tangent line at the point $x=2$ . Thus the slope of the tangent line at $x=2$ is 4. $\graph {f(x)=x^2,g(x)=4x-4}$

Find the slope of the secant line between $x = 2$ and $x = 7$ .

$\answer {9}$

Does this seem to be a good approximation for the slope of the tangent line at $x = 2$ ?

Yes No

Dylan thinks we can solve the problem by just picking something closer than 7. Find the slope of the secant line between $x = 2$ and $x = 3$ .

$\answer {5}$

Is this a good approximation for the slope of the tangent line at $x=2$ ?

Yes No

Is it better than the last attempt?

Yes No

Julia: Dylan, this still isn’t a great approximation...
Dylan: Well, I think we need to get even closer. Like, infinitesimally close! But how would we do that....
James: You guys need some help?
Julia and Dylan: James! How do we find the slope of a line at a point?
James: It isn’t too tough! Before, you were considering a certain point as your comparison. What if instead, you used the point you want to evaluate at plus something really small? Let’s call it $h$ .

How can you make the $h$ in $\frac {f(2+h)-f(2)}{(2+h)-2}$ approach 0?

Use $\lim _{h \to 0}$ . Use $\lim _{h \to \infty }$ . Divide the fraction by $h$ . Pick a function $f(x)$ so that $f(2)$ is 0.

Using the method you determined, approximate the slope of the tangent line at the point x=2.

$\answer {4}$

James: Want to know something really cool?
Julia and Dylan: What James?
James: The function we just discovered is how you determine a function’s derivative! Using that process, you can find the instantaneous rate of change at any point on a function!
Julia and Dylan: Wow! So cool!

On Your Own

Using what you’ve learned, find the derivative of the following functions at the given point.

Remember you can enter $\sqrt {x}$ as either

sqrt(...) or (...)^(1/2)

$g(x) = x^2+1$ , $x=2$

$\answer {4}$

$h(x) = \frac {1}{x}$ , $x=2$

$\answer {-0.25}$

$f(x)=3x^2+4x+2$ , $x=-1$

$\answer {-2}$

$f(x)=\sqrt {x^2+1}$ , $x=3$

$\answer {\frac {3}{\sqrt {10}}}$

$f(x) = x+x^{-1}$ , $x=4$

$\answer {15/16}$

By replacing the point in our formula for the derivative with $x$ , we may determine the derivative at any point on the function. Determine the derivative for the following functions.

$m(x) = x^3$

$\answer {3x^2}$

$n(x) = 3x+2$

$\answer {3}$

$f(x)=4-x^2$

$\answer {-2x}$

$f(x) = 12+7x$

$\answer {7}$

$f(x)=\frac {4}{x+1}$

$\answer {\frac {-4}{(x+1)^2}}$

In Summary

Julia: So why is it called a secant line?
James: It comes from the Latin word secare, which means ’to cut’.
Dylan: Ohh, I get it now! Because a secant line is a line that ’cuts’ a function!

In this lab we’ve (hopefully) learned the function for finding a derivative using the limit as $h$ approaches 0. We also learned what secant and tangent lines are. For your convenience, the important definitions are listed below.

A secant line is any line that connects any two points on a curve.

A tangent line is a line which intersects a differentiable curve at a point where the slope of the curve equals the slope of the line.

The derivative of $f(x)$ at $a$ is defined by the following limit: $\eval {\frac {d}{dx} f(x)}_{x=a} = \lim _{h\to 0} \frac {f(a+h) - f(a)}{h}.$