Implicit differentation

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It's easiest to use Implicit Differentiation if you realise that since both sides of the equation are equal, so are their derivatives.

So \frac{d}{dx}(3xy^2 + y^3) = \frac{d}{dx}(8)

Notice on the LHS you have a sum, and the derivative of a sum is the same as the sum of the derivatives. On the right you have a constant, and the derivative of a constant is 0.

So we have

\frac{d}{dx}(3xy^2) + \frac{d}{dx}(y^3) = 0.

The first part is a product, so we need to use the product rule. Since y is a function of x, we can find a derivative with respect to x, in terms of y. We do this using the chain rule.

Notice that \frac{dy}{dx}\cdot\frac{d}{dy} = \frac{d}{dx}. Also note that we are trying to FIND \frac{dy}{dx}.

So, using these pieces of information we have

3\left[x\frac{d}{dx}(y^2) + y^2\frac{d}{dx}(x)\right] + \frac{dy}{dx}\cdot\frac{d}{dy}(y^3) = 0

3\left[x\frac{dy}{dx}\cdot\frac{d}{dy}(y^2) + y^2\right] + 3y^2\frac{dy}{dx} = 0

3\left(2xy\frac{dy}{dx} + y^2\right) + 3y^2\frac{dy}{dx} = 0

6xy\frac{dy}{dx} + 3y^2 + 3y^2\frac{dy}{dx} = 0

\frac{dy}{dx}(6xy + 3y^2) = -3y^2

\frac{dy}{dx} = -\frac{3y^2}{6xy + 3y^2}.


Hope that helped. Have a go of the second one yourself.