Determine the equation of the tangent line.

Question

Here is a question in the textbook that I don't know how to do for my test!

Determine the equation of the tangent line to xy^2=sin(x+2y) at the point (0,0).

Answer 1

 

xy^2=sin(x+2y)

1y^2+2xy(dy/dx)=cos(x+2y)(1+2(dy/dx)   implicit differentiation

 

at this point you would normally solve for dy/dx but that is going to take some time  and the question only want to knwo the slope at 0,0. so just substitute in those values

 

0 + 0 =cos (0) (1+2(dy/dx))

0= 1(1+2(dy/dx)

 

dy/dx=-1/2

 

so the tangent line would be y =-1/2x