Question

Find the slope of the line tangent to x^3+2xy+y^2=49 at (1,6)

Answer 1

So find the derivative of the function at 1. Hopefully you know power rules

Answer 2

Woops just saw the y^2. My bad, totally goofed.

Answer 3

what do you do with the 2 in 2xy?

Answer 4

I guess, you simplify to a y= equation first, THEN take the derivative at 1.

Answer 5

So I believe you get f(x)=-x^3/2-2xy^1/2+49^1/2, then use power rules to find f'(1).

Answer 6

You know what, I could be completely wrong here. I'm not used to seeing equations in this form and finding derivatives of them, so hopefully a more experienced person will post :P Sorry x.x

Answer 7

thank you :)

Answer 8

Have you tried making it a y= equation?

Answer 9

Differentiate the function implicitly.

Answer 10

derive implicitly, get
x^3+2xy+y^2=49 at (1,6)
3x^2 + 2y + 2xy' + 2yy' = 0
2xy' + 2yy' = -(3x^2 + 2y)
y' (2x + 2y) = -(3x^2 + 2y)
y' = -(3x^2 + 2y)/(2x + 2y)
now subtitute x = 1 and y = 6 into the equation of y' above to get the slope