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Mathematics

MAT237Y1

Dan Dolderman

Fall

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Calc 1
Mon 9/22/08
3.2 - Derivatives!
0
Derivative: The slope of the tangent line. It’s all that stu▯ we did last week. Denoted f (x)
(pronounced f prime of x), it’s simply
0 f(x + h) ▯ f(x)
f (x) = lim
h!0 h
and is called the derivative with respect to x.
0
The domain of f (x) is all the x in the domain of f(x) AND where the above limit exists.
2
Example: Find the derivative with respect to x of f(x) = 3x ▯ 2.
Solution:
f (x) = limf(x + h) ▯ f(x)
h!0 h
2 2
[3(x + h) ▯ 2] ▯ [3x ▯ 2]
= lim
h!0 h
[3(x + 2xh + h ▯ 2] ▯ 3x + 2
= lim
h!0 h
3x + 6xh + 3h ▯ 2 ▯ 3x + 2
= lim
h!0 h
2
= lim6xh + 3h
h!0 h
= limh(6x + 3h)
h!0 h
= h!0m 6x + 3h
= 6x
Keep in mind that the derivative is a function, so you should be getting out something with x in
it. Only when we want to know what the derivative is at a speci▯c point (like, say, x = 3) do we
only get a number out. The derivative is the function that when you plug in a value for x, it spits
out the slope of f(x) at that value for x. Compare the graphs of f(x) = 3x ▯ 2 and f (x) = 6x.
1 Example: Let’s take a look at what the derivative of some line y = mx + b might be.
Solution:
0 f(x + h) ▯ f(x)
f (x) = lim
h!0 h
[m(x + h) + b] ▯ [mx + b]
= lim
h!0 h
mx + mh + b ▯ mx ▯ b
= lim
h!0 h
mh
= lim
h!0 h
= lim m
h!0
= m
So, the slope of the tangent line of a line is just the slope of the line itself! Well, that actually
kind of makes sense if we look at the graph of any ol’ line!
Now here’s a shocker: for f(t) a position function of a particle in rectilinear motion, the derivative
0
f (t) is the instantaneous velocity function and is (what else but)
0 f(t + h) ▯ f(t)
v(t) = f (t) = lim
h!0 h
Technically, it’s the instantaneous velocity function, but just simply call it the velocity function. If
we want average velocity, we’ll ask for the average velocity.
Example: Say a particle is moving by the position function f(t) = 4t + t, where f(t) is measured
in meters and t in seconds. Find the velocity function.
Solution:
[4(t + h) + (t + h)] ▯ [4t + t]
f (t) = lim
h!0 h
[4(t + 2th + h ) + t + h] ▯ 4t ▯ t
= lim
h!0 h
2 2 2
= lim 4t + 8th + 4h + t + h ▯ 4t ▯ t
h!0 h
2
8th + 4h + h
= h!0 h
h(8t + 4h + 1)
= h!0 h
= lim 8t + 4h + 1
h!0
= 8t + 1
Okay, so here I am talking about all this derivative stu▯, but I neglected to mention the fact that
the derivative doesn’t have to always exist. Remember when I ▯rst de▯ned it, I mentioned that the
domain of the derivative was the x’s in the

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