Expert: Sherry Wallin - 11/3/2009

Question
Use implicit differentiation to find the slope of the tangent line to the curve 4x^2+2xy–2y^3=8 at the point (-3,2).

I'm so lost...can you show how to solve this step by step? Thanks Sherry.

Answer
Sue~
    I'm not sure what part has you lost but to do implicit differentiation it is just like taking the derivative with respect to x but every time you have a y you need to multiply by y' and that is because of the chain rule: I moved the constant to the left so that f(x) = 0

f'(x)= (4x^2+2xy–2y^3-8)' = 2*4*x + 2*y*y' -3*2*y^2*y' - 0 = 0 gather all terms with a y' in them and factor the y' out: 8x + y'(2y-6y^2) = 0, move 8x to the right hand side: y'(2y-6y^2) = -8x divide both sides by (2y-6y^2): y' = -8x/(2y-6y^2), this is f'(x) expressed explicitly
The slope of the tangent line is the first derivative and at the point given is merely found by plugging in the point: -8(-3)/(2*2-6*2^2)= 24/-20 = -6/5-> this is the slope

Math Prof