Calculus problem, Find an equation of the tangent line to the graph of the function 2x^2-5xy+3y^3-32=0 at-1,2

((2x)^(2))-((5)(xy))+((3)(y)^(3))-(32)=0 at(-1,2)

Im having trouble figuring out why exactly I only place dy/dx for -5x and 9y^2

I'm assuming you did the differentiation correct and just wondering the reason why we have to put dy/dx every time we have to differentiate the y with respect to x. The reason for this is that:

you cannot differentiate 3y^3 with respect to x without implicit differentiation.

when you are differentiating 2x with respect to x,
d(2x)/dx = 2


but, you cannot differentiate 2y with respect to x unless you do the implicit differentiate.
example: You cannot have
d(2y)/dx ..i mean where's the x?


First you have to differentiate it with respect to y and then to x, and so

(d(2y) / dy ) (dy/dx) ...derive 2y with respect to y itself and then derive the derivation of 2y with respect to x. You can also imagine that dy looks like it is canceling out with each other.


I think the answer is y = 14/41x + 96/41 (I did not check my answer though!)

I hope that help!

You could either differentiate implicitly or you could take the partial derivatives.

1)
4x - 5 [y + xy' ] + 9y^2 y' =0
y' (-5x + 9y^2) = -4x + 5y
y'= -4x + 5y / (-5x + 9y^2)


2) fx = 4x - 5y
fy = -5x + 9y^2
dy/dx = -fx/fy = 5y -4x / 9y^2 -5x

Those both give you the same dy/dx for any point on the curve

-1,2 is not on the given curve => no tangent