Find the slope of the graph of the relation x^2y + 4y = 8 at the point (2, 1). I keep getting -1/8 and I'm really confused. Any help is accepted and greatly cherished.

Question

tkhunny · Answer

How do you manage that?  Please demonstrate your process.

kaleidoscopicsink · Answer

4y+x^2y --> 4 (dy/dy) + (y+ 2x * dy/dx) --> 4 (dy/dy) +2x(dy/dx) = -y --> -y / 4+2x   = dy/dx   **Which I know is horribly wrong

tkhunny · Answer

??  "Which I know is horribly wrong" -- Then why did you do it?

[x^2y] + 4y = 8

Let's take a closer look at the multiplication rule.

[x^2 (dy/dx) + 2xy] + 4 (dy/dx) = 0

kaleidoscopicsink · Answer

Okay, give me a second, I forgot about the product rule.

kaleidoscopicsink · Answer

Would the answer be -1/2 for the slope?

tkhunny · Answer

Again, how did you get that?

tkhunny · Answer

[x^2 (dy/dx) + 2xy] + 4 (dy/dx) = 0

(2,1)

[2^2 (dy/dx) + 2(2)(1)] + 4 (dy/dx) = 0

4 (dy/dx) + 4 + 4 (dy/dx) = 0

dy/dx = -4/8 = -1/2

Show your work in the future, please.  It is so much easier to redirect your from errors and encourage you with acceptable results.

kaleidoscopicsink · Answer

Using the product rule for x^2y you get: 2xy+x^2. Then going back to 4y being 4 --> 4 (dy/dx) + 2xy + x^2 * dy/dx --> -2xy / 4+x^2 then insert 1,2.

tkhunny · Answer

Pretty good, except for the tiny notation problem.

You mean -2xy / [4 + x^2].  You do NOT mean what you have written.

kaleidoscopicsink · Answer

Yep. cx Can you check if I did this right? 
If 5x^2 + y^4 = −9 then evaluate the second derivative of y with respect to x when x = 2 and y = 1
So 5x^2 goes to 10x and y^4 goes to 4y^3 then that derivative is -10x/4y^3, then take the derivative of that using the quotient rule which gives you -40y^3 +120xy^2 / 16y^6, finally insert the given vaalues and you get 12.5?
@tkhunny

tkhunny · Answer

Please post on another thread.  Sorry, gtg.

kaleidoscopicsink · Answer

It's okay. Have a great day. Thanks for the help. cx @tkhunny