Help on Equation of Tangent Q

[VCE1996]

Hi everyone,

I need help with this question.. if someone could please explain and provide solutions?

Find the equation of the tangent to the curve

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

at the point (-1,-1)

Thanks![/size][/size]

[kinslayer]

Use implicit differentiation with respect to x (treating y as a function of x):



Solve for dy/dx:



Evaluate dy/dx at the point (-1,-1):



The tangent line is the straight line with slope -5/7 passing through the point x = -1, y = -1:

[VCE1996]

Hey, thanks for the solutions you provided

Could you please put in the working out?

Ive been trying to figure out how you did the differentiation for an hour now literally.. I keep getting similar answers but a few terms are always wrong (positive negative or numbers etc) and Im getting more and more confused :/

[kinslayer]

I guess the only tricky part is the term. For this you need to use the product and chain rules.



The term is just using the chain rule.

[VCE1996]

I tried and got this:

-1+2xy+6y^2
        4x-y^2

[kinslayer]

Looks like you got all the derivatives correct, but you made a mistake when solving for dy/dx. Post your working, someone will be able to tell you where you went wrong.