Implicit differentiation product rule

Question

Whenever I look at the solution for the derivative of an implicit function, I see that the product rule is used for terms with two different variables. For example, for the equation $e^{xy^2}$=$x-y$ you have to solve for the derivative of $xy^2$ when taking the derivative of $e^{xy^2}$ and at that step you use the product rule. I'm confused because I thought that I could just treat $y^2$ as a constant in the term ${xy^2}$ and thus get $d(xy^2)/dx=y^2$. Why can't I do that?

Answer 1

For the reason that $y$ is a function of $x$; it is not a constant. The chain rule mandates you multiply by a factor of $y'$ when you differentiate the $y$.

Answer 2

What you have to understand is that $y$ is a function. You can think of it as $y(x)$. Thus, when you have $xy^2$, $y^2$ is actually a function, which is $[y(x)]^2$. The derivative of $xy^2$ would be:

$$(1)(y^2) + (x)(2)(y)(y'(x))$$ $$y^2 + 2xyy'(x)$$

It is not a constant. Sure, if you have something like $2y$, then the derivative is $2y'(x)$. Notice how if it was $2x$, then its just $2$. But since $y$ is a function , you must treat is as such.

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Usually, $y'(x)$, is just abbreviated as $y'$.

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Now that you know that, let's differentiate $e^{xy^2}$.

Thats $e^{xy^2}$ * the derivative of ${xy^2}$, which we calculated above. Thus, we have:

$$e^{xy^2} * [y^2 + 2xyy']$$

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Differentiating $x - y$ is easy. It's just $1 - y'$.

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In the end you have:

$$e^{xy^2} * [y^2 + 2xyy'] = 1 - y'$$

Answer 3

Here's a simple example:

$$y=x$$

Clearly, $\frac{dy}{dx}=1$, right? But if we treat $y$ as a constant, then you end up with $\frac{dy}{dx}=0$, since the derivative of a constant is always $0$. This clearly cannot be right, since $0\ne1$. Thus, it makes no sense, even in simple scenarios, to treat $y$ as a constant unless it actually is a constant.

Answer 4

I suppose that using the implicit function theorem could help since, up to the end, you do not need to worry about the fact that $y$ is a function of $x$.

Considering$$F=e^{xy^2}-x+y=0$$ $$F'_x=y^2 e^{x y^2}-1$$ $$F'_y=2 x y e^{x y^2}+1$$ $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{1-y^2 e^{x y^2}}{2 x y e^{x y^2}+1}$$