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)