Question

How do you find the slope of the tangent line to the curve `4xy^3+3xy=7` at the point (1,1) by using imlicit differentiation?

Answer 1

I found: Slope`=-7/15`

Explanation:

First you need to derive implicitly (rememberin that `y` will be a function of `x`) and get (using the Product Rule on the two terms on the left):
`4y^3+12xy^2(dy)/(dx)+3y+3x(dy)/(dx)=0`
collect `(dy)/(dx)`:
`(dy)/(dx)=(-4y^3-3y)/(12xy^2+3x)=`Slope
Now use the coordinates of your point:
`x=1`
`y=1`
`(dy)/(dx)=(-4-3)/(12+3)=-7/15`