Question

How do you use implicit differentiation to find `(dy)/(dx)` given `5x^3=-3xy+2`?

Answer 1

`-5x- y/x =y'`

Explanation:

`5x^3= - 3xy + 2` Differentiate the equation from the left and right

`15x^2(1)= -3xy'+ -3y` The product rule was used on -3xy

`15x^2+3y= -3xy'` Divide the -3x over to the other side

`-5x- y/x =y'` The answer.

For -3xy the product rule states f g' + f' g
Say -3x is f and y is g
-3x y' + -3 y

When you differentiate, especially implicit differentiation, you follow what the rules say, and this one is saying `dy/dx`.
a `dx/dx` causes a 1 (as seen when I did it on the `5x^3`), and a `dy/dx` will cause a y'. For other problems a `dx/dy` will cause a x' and the `dy/dy` will cause a 1.

Answer 2

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

Explanation:

`5x^3=-3xy+2`

Differentiate each term with respect to `x`

`d/(dx)(5x^3)=d/(dx)(-3xy)+d/(dx)(2)`

the first term on the RHS is differentiated using the product rule

`15x^2=-3(y+x(dy)/(dx))+0`

`15x^2=-3y-3x(dy)/(dx)`

rearrange for `(dy)/(dx)`

`(dy)/(dx)=(15x^2+3y)/(-3x)`

cancel the 3 out.

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