Find dy/dx by implicit differentiation. x^2 − 6xy + y^2 = 6

Answer 1

x² - 6xy + y² = 6

Differentiate both sides and get

2x - 6(y + xy') + 2yy' = 0

Now solve for y' and finish from here.

Answer 2

Why don't you just take the derivative of each term and then solve for dy/dx?

For example, if x^3 - 2xy + y^3 = 6, then taking the derivative gives

3x^2 - [ 2x(dy/dx) + 2y ] + 3y^2(dy/dx) = 0

3x^2 - 2x(dy/dx) - 2y + 3y^2(dy/dx) = 0

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

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

Answer 3

The derivative of x² (with respect to x) is 2x

The derivative of -6xy (with respect to x) requires the product rule since it is two things being multiplied together. The product rule says (u•v)' = u'•v + u•v' so if:

u = -6x

v = y

u' (with respect to x) = -6

v' (with respect to x) is dy/dx

So (-6xy)' = (-6)(y) + (-6x)(dy/dx)

The derivative of y² (with respect to x) is 2y•dy/dx (chain rule is needed because what's inside the function is not just plain "x")

The derivative of 6 = 0

Putting these together we have:

2x - 6y - 6x•dy/dx + 2y•dy/dx = 0

Then gathering all the dy/dx elements on the left side of the equation and everything else on the right side:

- 6x•dy/dx + 2y•dy/dx = 6y - 2x

Then, factoring out dy/dx we get:

dy/dx(-6x + 2y) = 6y - 2x

Then dividing both sides by (-6x + 2y) we get:

dy/dx = (6y - 2x)/(-6x + 2y) [then factor out a 2 from both the top and bottom of the right side to get:

dy/dx = 2(3y - x)/2(y - 3x) and canceling that 2 on top and bottom gives us our final answer:

dy/dx = (3y - x)/(y - 3x)