2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.


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


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:

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


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


In the end you have:

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

$\endgroup$
1
$\begingroup$

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 $y$.

$\endgroup$
2
  • $\begingroup$ How would I know when to treat it as a function and when to treat it as a constant? $\endgroup$ – user402571 Dec 29 '16 at 0:59
  • 1
    $\begingroup$ Context. If you are doing an implicit differentiation, the convention is that we use a $y$ for the dependent variable and $x$ for the independent one. However, you can use $dy/dx$ and $dx/dy$ notation to be explicit about which variable you think of as independent. $\endgroup$ – ncmathsadist Dec 29 '16 at 1:18
1
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ :'( Why do the downvotes hit me. $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 0:45
  • $\begingroup$ Oh, come on, at least leave me an explanation as to why this is a bad answer XD $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 0:48
  • $\begingroup$ If we treated y is a constant, wouldn't it result in 0 = 1? $\endgroup$ – Saketh Malyala Dec 29 '16 at 1:01
  • 1
    $\begingroup$ @SakethMalyala But $0\ne1$, thus, we cannot treat $y$ as a constant unless it is literally a constant function. $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 1:02
  • $\begingroup$ Of y=constant then x is also constant then dy/dx=0/0 $\endgroup$ – Fawad Dec 29 '16 at 3:15
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.