Question

Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10 at the point (5,1). The equation of this tangent line can be written in the form y=mx+b where m is:

Answer 1

I need to find m and B

Answer 2

So lets begin \[3y^2*y' + y' = 0\] I think that's how you simplify it.

Answer 3

Nope. Gonna have to restart.

Answer 4

wait, product rule?

Answer 5

So, let's start with the first term just so we can get the hang of what we need to do.

\(xy^{3}\). We have two variable expressions being multiplied, so we need to do product rule actually. Recall the product rule for derivatives is \(f'(x)g(x) + f(x)g'(x)\). So if we want to break this first one down, we can say:
\(f(x) = x\), \(f'(x) = 1\)
\(g(x) = y^{3}\), \(g'(x) = 3y^{2}*y'\)

Now, I only tack on the \(y'\) when I actually do a derivative operation to "y". So using product rule, the first term becomes:

\(y^{3} + x(3y^{2})y'\)

Does the derivative for that first term make sense?

Answer 6

okay, I believe so

Answer 7

Alrighty, cool. So how do you think the derivative of the 2nd term would turn out then? It's product rule again :)

Answer 8

\[(x) (y)*y' + (y)(1) * y'\]

Answer 9

scratch the last y

Answer 10

You only have a \(y'\) when you take the derivative of a y. Just because a term has a y in it doesn't mean you will have \(y'\) in it after you take the derivative. So if I do the derivative of xy.
\(f(x) = x\), \(f'(x) = 1\)
\(g(x) = y\), \(g'(x) = 1*y'\)

Only that one derivative will have a \(y'\) in it. So that means I would have:
\(y + x*y'\)
Make sure that makes sense to ya.

Answer 11

okay, now what do we do next

Answer 12

Do you see how I did that, though?

Answer 13

yes, implicit differentiation and product rule

Answer 14

okay, now what about those extra ys

Answer 15

okay so, it looks like we did something wrong here

Answer 16

When I try to enter the slop online, it says its incorrect.

Answer 17

I wrote it out incorrect as I was going through the steps. I missed a term, so that's on me :)

Answer 18

Okay, so I should have:
\(y^{3} + (3xy^{2})y' + y + (x)y'=0\)
\((3xy^{2})y' + (x)y' = -y-y^{3}\)
\(y'(3xy^{2} + x) = -y-y^{3}\)
\(y' = \frac{-y-y^{3}}{3xy^{2} + x}\)

Answer 19

okay, so we plug in "1" for y correct

Answer 20

and 5 for x :3

Answer 21

y' = -1/10

Answer 22

and y-1 = -(1/10)x + 1/2 correct?

Answer 23

yes

Answer 24

so how to make this in y = mx + b?

Answer 25

yes

Answer 26

\(y - 1 = \frac{-1}{10}(x-5)\)
\(y = \frac{-x}{10} + \frac{1}{2} + 1\)
\(y = \frac{-x}{10} + \frac{3}{2}\)

That should be your answer :P TO get into y = mx + b, you just solve for y using your point-slope form equation.

Answer 27

WE DID IT!!!!!

Answer 28

Thanks!

Answer 29

No problem :)

Answer 30

Thanks! as thx