I'm checking MIT maths course and I'm currently studying Implicit Differentiation and Inverse Functions (http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/). I don't understand one equation in "Worked Example". Why differentiation of squared y is equal to 2yy'? I can understand 2y but how this y' appears there?

Question

anonymous · Answer

Understanding this is important and, when you do it will become very clear. 
It's the chain rule (notice the DY/DY):
$$\frac{dy}{dy}\cdot \frac{dy}{dx} = \frac{dy}{dx}$$

So for example say
$$y^2$$

$$\frac{d}{dx}y^2 = \frac{d}{dy}y^2 \frac{dy}{dx}=2y \cdot \frac{dy}{dx}$$
So your left with a dy/dx in an equation that you can algebraically manipulate as an 'object or symbol', you will also notice that we are using 'taking a derivative' just like a normal operation such as addition or multiplication, that is we are doing it to both sides of the equation.
$$y^2=x$$
$$2y \cdot \frac{dy}{dx} = 1$$
Now divide each side by 2y and get
$$\frac{dy}{dx}= \frac{1}{2y}$$

Because the original equation stated y squared is equal to x we can gather that y is equal to the square root of x, and can replace y in the equation like
$$y^2=x;y=\sqrt{x}$$
So that the equation
$$\frac{dy}{dx} = \frac{1}{2y}= \frac{1}{2\sqrt{x}}$$