<p>Consider the curve x + xy + 2y^2 = 6. The slope of the line tangent to the curve at point (2,1) is</p>

<p>I got down to: y + x(dy/dx) + 2y(dy/dx) = -1 </p>

<p>How do you solve for dy/dx from here? Or is that even what you have to do?</p>

<p>Consider the curve x + xy + 2y^2 = 6. The slope of the line tangent to the curve at point (2,1) is</p>

<p>I got down to: y + x(dy/dx) + 2y(dy/dx) = -1 </p>

<p>How do you solve for dy/dx from here? Or is that even what you have to do?</p>

<p>well you can always subtract y, factor out dy/dx, and then divide y-1 by (x+2y) - but i'm pretty positive that's not what you do.</p>

<p>If you are taking the derivative of 2y^2 in terms of x, it's just =0 because 2y^2 is essentially just a constant.</p>

<p>anyway, your derivative should be</p>

<p>1+1(dy/dx)+0=6

so dy/dx =5</p>

<p>The answer to the question is -1/3, I forgot how to do this whole tangent line thing.</p>

<p>what is the derivative of 2y^2? Is it 2(2)(y)(2y)(dy/dx) or 8y^2(dy/dx)?</p>

<p>in respect to x (d/dx), which is what the problem is asking, i believe its 2y(dy/dx), chain rule.</p>

<p>Okay i got it. The original is x + xy + 2y^2 = 6.</p>

<p>the derivative of x = 1.

the derivative of xy= (dy/dx)(x)+y

the derivative of 2y^2= 2(2y)(dy/dx) = 4y(dy/dx)</p>

<p>Add them up: 1+x(dy/dx)+y+4y(dy/dx) = 0

x(dy/dx)+4y(dy/dx)= -1-y

(dy/dx) = -1-y / (x+4y)</p>

<p>Then you substitute in the points to get -1/3 as the answer.</p>

<p>thank you!!!</p>

<p>johntam's right, good job.</p>

<p>hey here is another problem:

x^2 + 2xy - 3y = 3, then the value of dy/dx at x=2 is?

answer is 2</p>

<p>take derivative of the problem:</p>

<p>2x + (2y + (2x(dy/dx)) - 3 dy/dx = 0

2x(dy/dx) - 3 dy/dx = -2x - 2y

-2x - 2y / 2x - 3 = dy/dx

x=2

-4 - 2y / 4 -3 = dy/dx

I'm stuck here, the answer is -2</p>

<p>Thanks!</p>

<p>lol i feel dumb, i figured it out, you have to plug in 2 to the original equation to get the y value :/</p>