slope of the tangent line?

i agree with the above answer that, as written, this problem can't be solved. but i do assume that this is a typical calculus problem where the expression given is equal to a constant. to solve for the slope, the first derivative must be found:

d/dx (x^3+2xy+y^2) = d/dx (constant)

3x^2+2y+2x*dy/dx+2y*dy/dx = 0

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

plug in x=1 and y=1:

dy/dx = (-3-2)/(2+2) = -5/4

using (1,1) and y=mx+b:

1 = (-5/4)(1) + b

b = 9/4

thus the equation of the line is y = -5/4x + 9/4

x^3 + 2xy + y^2 = 4

Take the derivative of each side with respect to x.

You will have

3x^2 + 2y + 2xy' + 2yy' = 0

Move the terms with y' to the right side.

3x^2 + 2y = -2xy' - 2yy'

Factor out y' on the right side

3x^2 + 2y = y'(-2x - 2y)

Divide both sides by -2x-2y

y' = (3x^2 +2y)/(-2x-2y)

Now plug in your values for the point (1,1). That is, x =1 and y =1

y' = (3+2)/(-2-2)

y' = 5/-4

y' = -5/4

This is your slope.

Now use the equation

y - y1 = m(x-x1), where m is the slope and (x1,y1) is the given point.

y - 1 = (-5/4)(x-1)

Either something is missing, or one of the "plus" signs in your expression should be an equals sign. As written,, "x^3 + 2xy + y^2" is not an equation (it's missing an equals sign).

The question cannot be answered as written.

if i remember properly all you're able to desire to do is discover the deverative. as quickly as you do, in simple terms plug in 2 and thats the slope as for a manner....i kinda forgot, its been a year, sorry all i remember is that e^x is chain rule, and ln x is a million/x