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JOEL LEWIS: Hi.

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Welcome to recitation.

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In lecture you've learned how
to compute derivatives of

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polynomials, and you've learned
the relationship

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between derivatives and
tangents lines.

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So let's do a quick
example that puts

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those two ideas together.

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So here I have a question
on the board--

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compute the tangent line to
the curve y equals x cubed

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minus x at the point 2, 6.

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So why don't you take a minute,
work on that yourself,

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pause the video, we'll come back
and we'll do it together.

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All right.

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Welcome back.

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So we have this function, y
equals x cubed minus x.

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Let's just draw a quick
sketch of it.

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So looks to me like it has
0's at 0, 1, minus 1.

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And it sort of does something
like this in between.

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Very rough sketch there.

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And way over-- well, OK.

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We'll call that the
point, 2, 6.

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Feel a little sketchy,
but all right.

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OK.

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So we want to know what the
tangent line to the curve at

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that point is.

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So in order to do that, we
need to know what its

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derivative is, and then that'll
give us the slope.

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And then with the slope, we have
the slope and we have a

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point, so we can slap that into,
say, your point-slope

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formula for a line.

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So, all right, so the
derivative of this

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function is y prime.

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So, OK, so here we have a sum
of two things, and they're

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both powers of x.

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And so we learned our rules
for a power of x that the

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derivative of x to the n is n
times x to the n minus 1.

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And so we also learned that the
derivative of a sum of two

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things is the sum of
the derivatives.

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So in this case, so the
derivative of x cubed minus x

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is 3x squared minus 1.

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OK.

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So this is the slope of the
function in terms of x.

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But in order to compute the
tangent line, we need the

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slope at the particular
point in question.

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Right?

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This is really important.

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So we aren't going to use 3x
squared plus 1 as our slope.

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Right?

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We want the slope at the
point, x equals 2.

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Right?

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OK.

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So what we want for the slope
of the tangent line

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is y prime of 2.

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Right?

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We want it at this point
when x is equal to 2.

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So that's equal to, well,
3 times 2 squared is

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12 minus 1 is 11.

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So this is the slope, this is
the slope of the tangent line.

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I just want to say this one
more time for emphasis.

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This is a really common mistake
that we see on lots of

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homework and exams when
teaching calculus.

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You have to remember that when
you compute the slope of the

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tangent line, you compute the
derivative and then you need

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to plug in the value of x at
the point in question.

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Or the value of x and y.

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You know, you need to plug in
the values of the point that

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you have. So here, the
derivative is 3x squared minus

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1, so the slope at the
point, 2, 6 is 11.

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It's just a number.

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The slope at that point is
that particular number.

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OK, so now, to compute the
tangent line, we have a point,

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2, 6 and we have a slope, 11.

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So we can plug into
point-slope form.

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So the equation of
the-- one the--

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tangent line is y minus 6 is
equal to 11 times x minus 2.

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Right?

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So it's y minus y0 is equal to
the slope times x minus x0.

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If you like, some people
prefer to write their

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equations of their lines in
slope-intercept form.

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So if you wanted to do that,
you could just multiply

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through by 11 and then bring
the constants together.

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So we could say, or,
y equals 11x--

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well, we get minus 22 plus 6--

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is minus 16.

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So either of those is
a perfectly good

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answer to the question.

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So that's that.

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