MHB Question about implicit differentiation

[find_the_fun]

Say you want to find the slop of a tangent line of the circle [math]x^2+y^2=25[/math]

I was following the directions here. I don't completely understand how the derivative of [math]y^2[/math] becomes [math]2y\frac{dy}{dx}[/math]. Shouldn't it become 0 if we are taking the derivative with respect to [math]x[/math]? The website explains

Recall that the derivative (D) of a function of x squared, (f(x))² , can be found using the chain rule

but to me that's not really saying anything; while I can see they used the chain rule why DID they use the chain rule, it seems like they just pulled it out of thin air.

 

[Euge]

Say you want to find the slop of a tangent line of the circle [math]x^2+y^2=25[/math]

I was following the directions here. I don't completely understand how the derivative of [math]y^2[/math] becomes [math]2y\frac{dy}{dx}[/math]. Shouldn't it become 0 if we are taking the derivative with respect to [math]x[/math]? The website explains but to me that's not really saying anything; while I can see they used the chain rule why DID they use the chain rule, it seems like they just pulled it out of thin air.

Since $y^2$ varies with $x$, you can expect $\frac{d(y^2)}{dx}$ to be nonzero. They computed $\frac{d(y^2)}{dx}$ by the chain rule:

$$ \frac{d(y^2)}{dx} = \frac{d(y^2)}{dy} \frac{dy}{dx} = 2y \frac{dy}{dx}$$.

In problems involving implicit differentiation, you are usually given an equation of the form $F(x,y) = 0$ where $y$ depends on $x$, and you are asked to find $\frac{dy}{dx}$. If $F(x,y) = 0$ can't be solved explicitly for $y$, it is crucial to use the chain rule to find $\frac{dy}{dx}$. For example, consider the equation

$$xe^y + y = 0$$.

This equation can't be solved explicitly for $y$, but we can find $\frac{dy}{dx}$ as follows:

$\displaystyle \frac{d}{dx}(xe^y + y) = \frac{d}{dx}(0)$,

$\displaystyle \frac{d}{dx}(xe^y) + \frac{dy}{dx} = 0$,

$\displaystyle e^y + xe^y \frac{dy}{dx} + \frac{dy}{dx} = 0$,

$\displaystyle (xe^y + 1)\frac{dy}{dx} = -e^y$,

$\displaystyle \frac{dy}{dx} = -\frac{e^y}{xe^y+ 1}$.

Hence, we know the slope of the tangent line to the curve $xe^y + y = 0$ at the origin; it is

$\displaystyle \frac{dy}{dx}|_{(x,y) = (0,0)} = -1$.

 

[find_the_fun]

I guess what I'm saying is for me the equality

...
$$ \frac{d(y^2)}{dx} = \frac{d(y^2)}{dy} \frac{dy}{dx}$$.
...

is not apparent. Can you please explain how $$ \frac{d(y^2)}{dx} = \frac{d(y^2)}{dy} \frac{dy}{dx}$$?

 

[Euge]

I guess what I'm saying is for me the equality

is not apparent. Can you please explain how $$ \frac{d(y^2)}{dx} = \frac{d(y^2)}{dy} \frac{dy}{dx}$$?

Let $y = u(x)$ and $f(u) = u^2$. Then $y^2= f(u(x))$. So by the chain rule,

$$ \frac{d}{dy}(y^2) = \frac{d}{du}(f(u)) \frac{du}{dx} = 2u(x) \frac{du}{dx} = 2y \frac{dy}{dx}$$.

 

[find_the_fun]

I finally got it. Here's how I understood it:
We know [math]y[/math] is a function of [math]x[/math], we just don't know what that function is. So [math]y^2[/math] is a composite function and that's why we use the chain rule.
[math]\frac{d}{dx}(y^2)=\frac{d}{dx}((f(x))^2[/math] and by the chain rule that becomes [math]frac{d}{dx}(x^2) \cdot \frac{dy}{dx}[/math]