Class Notes (837,550)
Canada (510,314)
Mathematics (2,843)
MAT237Y1 (53)
Lecture

3.2 derivatives.pdf

4 Pages
60 Views
Unlock Document

Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
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
More Less

Related notes for MAT237Y1

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit