A curve C has the equation x^3 + 2xy- x - y^3 -20 = 0. Find dy/dx in terms of x and y.
Every term needs to be differentiated in terms of x:
For the x3 term, differentiated it becomes 3x2 (multiply the whole term by the initial power, 3, and the reduce the power by 1 and replace it, 3 therefore becomes 2)
For the 2xy term, the chain rule must be used here (u'v + uv', where u' is u differentiated and v' is v differentiated). so set 2x = u, therefore u' = 2, and set v = y, therefore v' = dy/dx as there are no x terms present and y differentiated with respect to x. This means that the final term is 2(y) + 2x(dy/dx).
For the -x term, x differentiated just becomes the integer "in front of x", which is -1 in this case (for 3x it would become 3, for -100x it would become -100)
For the -y3 term, this would follow the same rule as the x3 term, but as it is y term being differenentiated we simply need to include a dy/dx term to it: -3y2 dy/dx
For the -20 term, as this is just a number, it differentiates to 0. This is the same for the 0 term on the other side of the equals sign; it remains as 0.
This leaves us with the equation: 3x2 + 2y + 2x(dy/dx) -1 -3y2 dy/dx = 0
As we are asked to find dy/dx in terms of x and y, this means the format should be dy/dx = something in x and y.
The easiest way to do this is put all the dy/dx on one side to start:
3x2 + 2y -1 = (3y2 )(dy/dx) - (2x)(dy/dx) which more simply becomes: 3x2 + 2y -1 = (3y2 - 2x)(dy/dx)
Finally, both sides can be divided by (3y2 - 2x) to give dy/dx = (3x2 + 2y -1)/(3y2 - 2x)
