find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Advice please if i am close

Cheers

When you have \(\displaystyle f(y)\), it is the same as evaluating \(\displaystyle f(x)\) except that there is a \(\displaystyle \frac{dy}{dx}\) attached (multiplied).

So deriving \(\displaystyle y^2 = x\) gives

\(\displaystyle 2y\frac{dy}{dx} = 1\)

\(\displaystyle \frac{dy}{dx} = \frac{1}{2y}\)

This is simply a special case of the chain rule: notice that when evaluating x's this rule is hidden by the fact \(\displaystyle \frac{dx}{dx} = 1\).

Product rules with \(\displaystyle f(y)g(x)\) work as you would expect. Can you use this technique to answer your question?