Yes we are studying implicit differentiation and I did try using it but since we covered "simpler" examples in classes and we are assigned tougher problems to practice at home I don't even know how to set this problem properly from the start.
I tried starting it like a somewhat similar problem I found on another site but the answers I get keep coming up wrong. This is one way another site had as a possible solution:
d(y^2)/dt + d(x)/dt = 0
y^2(dx/dt) + x(dy/dt) + dx/dt = 0
Then I tried to solve it from there. This probably doesn't make any sense but it was the only other help I could find online to a problem similar to mine.
The underlying thought here is that both y and x are functions of t. Of course that means that y^2 is also a function of t, as is y^2 + x. And we apply the chain rule.
I am going to go step by step so that you see what is going on.
\(\displaystyle y^2 + x = 3 \implies \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(3) = 0.\) Make sense?
\(\displaystyle But\ \dfrac{d}{dt}\left(y^2 + x\right) = \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = \dfrac{d}{dt}(y^2) + \dfrac{dx}{dt}.\) Addition rule, right?
\(\displaystyle Now \dfrac{d}{dt}(y^2) = \dfrac{d}{dy}(y^2) * \dfrac{d}{dt}(y).\) Chain rule.
\(\displaystyle But\ \dfrac{d}{dy}(y^2) = 2y.\) Power rule.
\(\displaystyle And\ \dfrac{d}{dt}(y) = \dfrac{dy}{dt}.\)
\(\displaystyle So\ \dfrac{d}{dy}(y^2) = 2y * \dfrac{dy}{dt}.\) Substituting.
\(\displaystyle Giving\ \dfrac{d}{dt}(y^2) + \dfrac{d}{dt}(x) = 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt}.\) More substitution.
\(\displaystyle Thus,\ 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.\) And more substitution.
Implicit differentiation just skips a bunch of steps and really is a result of the chain rule.
\(\displaystyle y^2 + x = 3 \implies 2y * \dfrac{dy}{dt} + \dfrac{dx}{dt} = 0.\) Proceed.
