JOEL LEWIS: Hi.
Welcome to recitation.
In lecture you've learned how to compute derivatives of
polynomials, and you've learned the relationship
between derivatives and tangents lines.
So let's do a quick example that puts
those two ideas together.
So here I have a question on the board--
compute the tangent line to the curve y equals x cubed
minus x at the point 2, 6.
So why don't you take a minute, work on that yourself,
pause the video, we'll come back and we'll do it together.
All right.
Welcome back.
So we have this function, y equals x cubed minus x.
Let's just draw a quick sketch of it.
So looks to me like it has 0's at 0, 1, minus 1.
And it sort of does something like this in between.
Very rough sketch there.
And way over-- well, OK.
We'll call that the point, 2, 6.
Feel a little sketchy, but all right.
OK.
So we want to know what the tangent line to the curve at
that point is.
So in order to do that, we need to know what its
derivative is, and then that'll give us the slope.
And then with the slope, we have the slope and we have a
point, so we can slap that into, say, your point-slope
formula for a line.
So, all right, so the derivative of this
function is y prime.
So, OK, so here we have a sum of two things, and they're
both powers of x.
And so we learned our rules for a power of x that the
derivative of x to the n is n times x to the n minus 1.
And so we also learned that the derivative of a sum of two
things is the sum of the derivatives.
So in this case, so the derivative of x cubed minus x
is 3x squared minus 1.
OK.
So this is the slope of the function in terms of x.
But in order to compute the tangent line, we need the
slope at the particular point in question.
Right?
This is really important.
So we aren't going to use 3x squared plus 1 as our slope.
Right?
We want the slope at the point, x equals 2.
Right?
OK.
So what we want for the slope of the tangent line
is y prime of 2.
Right?
We want it at this point when x is equal to 2.
So that's equal to, well, 3 times 2 squared is
12 minus 1 is 11.
So this is the slope, this is the slope of the tangent line.
I just want to say this one more time for emphasis.
This is a really common mistake that we see on lots of
homework and exams when teaching calculus.
You have to remember that when you compute the slope of the
tangent line, you compute the derivative and then you need
to plug in the value of x at the point in question.
Or the value of x and y.
You know, you need to plug in the values of the point that
you have. So here, the derivative is 3x squared minus
1, so the slope at the point, 2, 6 is 11.
It's just a number.
The slope at that point is that particular number.
OK, so now, to compute the tangent line, we have a point,
2, 6 and we have a slope, 11.
So we can plug into point-slope form.
So the equation of the-- one the--
tangent line is y minus 6 is equal to 11 times x minus 2.
Right?
So it's y minus y0 is equal to the slope times x minus x0.
If you like, some people prefer to write their
equations of their lines in slope-intercept form.
So if you wanted to do that, you could just multiply
through by 11 and then bring the constants together.
So we could say, or, y equals 11x--
well, we get minus 22 plus 6--
is minus 16.
So either of those is a perfectly good
answer to the question.
So that's that.