Question

Use implicit differentiation to find the slope of the tangent line to the curve

4 x^2 + 4 xy + 2 y^3 = -80
at the point ( -2,-4 ).

Answer 1

What have you tried in this problem so far?

Answer 2

Honestly I'm lost lol but i'm open to any suggestions.

Answer 3

Alright.
The tried and true start is often just taking the derivative of both sides of the equation with respect to x.
Our intent is to find an equation between dy/dx (the slope we need), x and y (the point we have).

Answer 4

\( \displaystyle \frac{d}{dx} \left( 4x^2 + 4xy + 2y^3 \right) = \frac{d}{dx} \left( -80 \right) \)
Like this, and we take the derivative of each term taking care of y's with chain rule.

Answer 5

May I see the following steps after that?

Answer 6

I'm so sorry I'm with a friend trying to solve this and we're lost. Your help means a lot. So thank you.

Answer 7

Differentiation by terms (and last derivative of a constant is 0)
\( \displaystyle \frac{d}{dx} \left( 4x^2 \right) + \frac{d}{dx} \left( 4xy \right) + \frac{d}{dx} \left( 2y^3 \right) = 0 \)
The first term requires power rule. I can go over it if needed, but that is the easiest to convert.
The second term is a product, so we need product rule (pulling out a constant as well:
\( \displaystyle \frac{d}{dx} \left( 4xy \right) = 4 \left( \frac{d}{dx} \left( x \right) + \frac{d}{dx} \left( y \right) \right) \)
We can leave the dy/dx as is because we need it later to be there. The dx/dx is just equal to 1. So overall you have 4 + 4 dy/dx.
The last term is strictly a function of y, so chain rule is in order:
\( \displaystyle \frac{d}{dx} \left( 2y^3 \right) = \frac{d y^3}{dy} \times \frac{dy}{dx} \)
dy^3/dy is a derivative of y^3 with respect to y now, so we use power rule now.

Should I go further or can you figure this out now?

Answer 8

If you can go further that would be great. I'm taking note of this.

Answer 9

Got it.
So, I make the proper updates from above to our equation.
\( \displaystyle 8x + 4 + 4 \frac{dy}{dx} + 6y^2 \frac{dy}{dx} = 0 \)
(Power rule on 4x^2: d/dx(4x^2) = 4 *2x^(2-1) = 8x)
(Power rule on 2y^3: d/dy (2y^3) = 2*3y^(3-1) = 6y^2)

Now we have an equation involving dy/dx, x, and y.
The remaining step is using our point, (x,y) = (-2,-4), to remove the variables and solve for dy/dx.

Answer 10

You might also be able to solve for dy/dx at that point, but it does not matter in this case what order is taken to solve for dy/dx. I simply like dealing with numbers more than dealing with variables. :)

\( \displaystyle 8(-2) + 4 + 4 \frac{dy}{dx} + 6(-4)^2 \frac{dy}{dx} = 0 \)
\( \displaystyle -16 - 4 + 4 \frac{dy}{dx} + 96 \frac{dy}{dx} = 0\)
Here on is simply solving for dy/dx. This is the slope of the tangent line at that point!

Answer 11

Oh man thank you so much. Its been a frustrating problem for me. Thanks dude you're the best.

Answer 12

You're welcome, glad to help out! :)
If there is anything I need to clarify here, I am happy to elaborate!

Answer 13

Thank again!!!!!

Answer 14

you* haha