Question

How do you find `dy/dx` by implicit differentiation given `y^2+3y-5=x`?

Answer 1

`dy/dx = 1 /(2y+3)`

Explanation:

`y^2+3y−5=x`

Differentiate wrt `x` and we get:

`d/dx(y^2) + d/dx(3y) − d/dx(5) = d/dx (x)`
`:. d/dx(y^2) + 3dy/dx − 0 = 1`
`:. d/dx(y^2) + 3dy/dx = 1`

This is where we get stuck because we can differentiate `y` wrt `x` to get `dy/dx` but we cannot different `y^2` wrt `x`. We can however use the chain rule, to get:

`dy/dx*d/dy(y^2) + 3dy/dx = 1`

And we can now handle this:

`dy/dx*2y + 3dy/dx = 1`
`:. 2ydy/dx + 3dy/dx = 1`
`:. (2y+3)dy/dx = 1`
`:. dy/dx = 1 /(2y+3)`

we can skip the direct application of the chain rule and use the result it yields which is that if we differentiate a function of `y` wrt `x` then we get `dy/dx` times the function of `y` differentiated wrt `y` - It is this that is known as "Implicit Differentiation".