Differentiation involving implicit and parametric.

Question

I was asked to differentiate the term,

$$X^3 + XY^2 - Y^3$$

For such I reached,

$$3X^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$

The apparent answer is,

$$3X^2 + Y^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$

How exactly this is reached as I am new to the whole implicit and parametric function.

Answer 1

You forgot about the product rule. Assuming $y=y(x),$

\begin{align*} \frac{d}{dx} \left( x^3 +xy^2 - y^3 \right) &= 3x^2 + \frac{d}{dx} \left(xy^2 \right) -3y^2 \frac{dy}{dx} \\ &= 3x^2 + 2xy \frac{dy}{dx} + y^2 - 3y^2 \frac{dy}{dx} \end{align*}

Answer 2

Chain rule.

If $y = y(x)$ then $\frac{\text{d}}{\text{d}x}y(x) = \frac{\text{d}y(x)}{\text{d}y}\frac{\text{d}y}{\text{d}x}$.

Hence for example:

$$\frac{\text{d}y^3}{\text{d}x} = \frac{\text{d}y^3}{\text{d}y}\frac{\text{d}y}{\text{d}x} = 3y^2\frac{\text{d}y}{\text{d}x}$$

Answer 3

Let $$f(X,Y)=X^3 + XY^2 - Y^3$$ Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. \begin{align} \frac{d}{dX}(X^3 + XY^2 - Y^3)&=\frac{d}{dX}(X^3)+\frac{d}{dX}(XY^2)-\frac{d}{dX}(Y^3)\\ &=3X^2+\left(X\frac{d}{dX}(Y^2)+Y^2\frac{d}{dX}(X)\right)-\frac{d(Y^3)}{dY}\frac{dY}{dX}\\ &=3X^2+\left(X\frac{d(Y^2)}{dY}\frac{dY}{dX}+Y^2\right)-3Y^2\frac{dY}{dX}\\ &=3X^2+\left(2XY\frac{dY}{dX}+Y^2\right)-3Y^2\frac{dY}{dX}\\ &=3X^2+Y^2+\left(2XY-3Y^2\right)\frac{dY}{dX}\\ \end{align} Where we have used the product rule and the chain rule for the middle term in brackets when differentiating terms with both $X$ and $Y$ involved, and the chain rule for the last derivative with $Y$ terms alone. Since we differentiate W.R.T. $X$ we convert $\frac{d(g(Y))}{dX}$ to $\frac{d(g(Y))}{dY}\frac{dY}{dX}$ using the chain rule to then allow us to differentiate the function in $Y$, $g(Y)$ W.R.T. $Y$. We then end up with an explicit derivative in terms of $X$ and $Y$.