Question

Question #6bfa5

Answer 1

(a) `dy/dx=-(3x^2+5y)/(5x+3y^2)`
(b) `y=-37/27x-19/9`

Explanation:

(a) Differentiate both sides of the equation.
Take notice that `(d)/(dx)y^3` = `d/(dy)y^3*((dy)/(dx))=3y^2(dy)/(dx)` (Chain Rule)
and `d/(dx)xy` = `y+x(dy/dx)` (Product Rule).

`d/(dx)(x^3+y^3+5xy)=d/dx65`
`3x^2+3y^2(dy)/(dx) + 5(y+x(dy)/dx)=0`
`(5x+3y^2)dy/dx=-3x^2-5y`

`dy/dx=-(3x^2+5y)/(5x+3y^2)`.

(b) Substitute `x=3,y=2` to `dy/dx=-(3x^2+5y)/(5x+3y^2)`.
The slope of the tangent line is `dy/dx=-(3*3^2+5*2)/(5*3+3*2^2)=-37/27`.

Therefore, the tangent line is:
`y-2=-37/27(x-3)`
`y=-37/27x-19/9`.