Question

How do you use implicit differentiation to find `(dy)/(dx)` given `3x^2+3=ln5xy^2`?

Answer 1

`(dy)/(dx) = y(3x - 1/(2x))`

Explanation:

Implicit differentiation is no different to normal, you just have to make sure to apply the chain rule when differentiating the `y` terms, ie `y` will differentiate to `(dy)/(dx)`.

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

When we differentiate the log I used a combination of chain and product rules, but you could also rewrite it as :

`ln(5xy^2) = ln(5) + ln(x) + ln(y^2)`

using rules of logs and then just use chain rule.

`6x = 1/(5y^2x)*(5y^2 + 10xy(dy)/(dx))`

`30x^2y^2 = 5y^2 + 10xy(dy)/(dx)`

`5y^2(6x^2 - 1) = 10xy(dy)/(dx)`

`y/2(6x - 1/x) = (dy)/(dx)`

`therefore (dy)/(dx) = y(3x - 1/(2x))`