Taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$ I am a bit confused about taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$.
$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2 y^2)}{dx} $$
Edit: Incorrect step
$$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 10xy^2 + 5x^22y\frac{dy}{dx}$$
$$=  3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 5 \left(2xy^2 + x^22y\frac{dy}{dx}\right)$$
...
So I don't get how we got $2y\frac{dy}{dx}$ from $y^2$. Why it is not $2\frac{dy}{dx}$???
 A: Well, basically it is simple chain rule. So, what is $(y^{2})'$? It is simply $$(f(x)^{2})'$$ which looks like simple chain rule. Just derivate the entire function to get $2f(x)$ and multiply with the inner function's derivative which is $f'(x)$. So you get
$$2f(x)f'(x)$$
Which is simply
$$2y\frac{dy}{dx}$$
A: The 2nd line should be:
$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2y^2)}{dx}$$
I think this is the key to why you are not understanding how to implicitly differentiate that term.
When you implicitly differentiate, you differentiate both sides of the equation with respect to $x$ (or whatever variable suits you). This is "doing" $\frac{d}{dx}$ to both sides.
To implicitly differentiate $x=y^2$, for example, you do this:
Let: $x=y^2$
$$\frac{d(x)}{dx}=\frac{d(y^2)}{dx}$$
$$1=1\times\frac{d(y^2)}{dx}$$
$$1=\frac{dy}{dy}\times\frac{d(y^2)}{dx}$$
Here, you are allowed to swap the bottoms of the fractions, just like if you had $\frac{1}{3}\times\frac{4}{2}$.
$$1=\frac{dy}{dx}\times\frac{d(y^2)}{dy}$$
Now you can differentiate $y^2$ like you "normally" do as you are differentiating with respect to the same variable, $y$.
$$1=2y\frac{dy}{dx}$$
The key to implicit differentiation is that you can only differentiate (as in doing what you normally do - ie. $x^2$ goes to $2x$ when differentiated) a function containing a variable with respect to that variable. To solve the problem of differentiating a function with $y$s in with respect to $x$, it may help to multiply the fraction by "1" (ie. $\frac{dy}{dy}$) as shown above.
So: $$\frac{d(x^2)}{dx}=2x$$
But:
$$\frac{d(y^2)}{dx} ≠ 2x$$
Normally, when you differentiate a function like $y=x^3$, you're still "doing the same thing to both sides":
$$y=x^3$$
$$\frac{d(y)}{dx}=\frac{d(x^3)}{dx}$$
$$\frac{dy}{dx}=3x^2$$
If the thing you are trying to differentiate has more than one kind of variable in it, like $xy$, then you can still use the normal differentiation rules:
Eg. Let $w=xy$. Find $\frac{dw}{dx}$.
$$w=xy$$
$$\frac{dw}{dx}=\frac{d(xy)}{dx}$$
You can use the product rule in the same way as before by letting $u = x$ and $v = y$.
Normally, $z=uv \implies z'=u'v+uv'$.
So: $$\frac{dw}{dx}=\frac{d(xy)}{dx}=x'y + xy'$$
$$\frac{dw}{dx}=\frac{dx}{dx}y + x\frac{dy}{dx}$$
$$\frac{dw}{dx}=y + x\frac{dy}{dx}$$
A: I hate the way implicit derivatives are taught, but rather than boring you with my own method, the short answer is this - The derivative of $x^2$ is $2x\frac{dx}{dx}$.  For the same reason, the derivative of $y^2$ is $2y\frac{dy}{dx}$.  It's just that on the first one, since $\frac{dx}{dx}$ is 1, it goes away, leaving you with only $2x$.
A: An expression $f(x,y)=g(x,y)$ is implicitly differentiated by
$$ \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial x} x' =\frac{\partial g}{\partial y} y' + \frac{\partial g}{\partial x} x'$$
but with $x'=1$ in your case.
What I get is
$$ \left(6 y (x^2+y^2)^2 \right) y' + \left( 6 x (x^2 + y^2)^2 \right) 1 = \left( 10 x^2 y \right) y' + \left(10 x y^2 \right) 1$$
which needs some simplifications.
The specific question as to why $\frac{{\rm d} }{{\rm d} x}y^2 = 2 y y'$ use the rule above to get
$$ \frac{\partial y^2}{\partial y} y' + \frac{\partial y^2}{\partial x} 1 = (2 y) y' + (0) 1 = 2 y y' $$
