Given the Differential equation $y'=-2xy^{2}$.

Find the derivative $\frac{d(y')}{dx}$!

My approach, which is not correct according to Wolfram Alpha:

1. Plugging in: $\frac{d(y')}{dx}=\frac{d(-2xy^{2})}{dx}$
2. Pulling out constants #1:$-2\cdot\frac{d(x^{1}\cdot y^{2})}{dx}$
3. Pulling out constants #2:$-2\cdot y^{2}\cdot\frac{d(x^{1})}{dx}$
4. Do the differential: $-2\cdot y^{2}\cdot1=\frac{d(y')}{dx}$

Wolfram Alpha computes this: $\frac{d(y')}{dx}=-4x\cdot y\cdot y'(x)-2y^{2}$

Since i do not have the Pro-Version of Wolfram Alpha and i am curious of knowing the maths behind it: What steps happen here and why can't i do it in my way ?

Background info:

I want to find those Derivatives in order to compute a numerical approximation for the Differential equation using a Taylor series expansion of 4th order.

Edits (after comments and answers were given):

So by the product rule, stated as follows:

$\left(f(x)\cdot g(x)\right)^{'}=f'(x)\cdot g(x)+f(x)\cdot g'(x)$

I identified the following terms as the elements of that product rule:

• $f(x)=x$
• $g(x)=\left(y(x)\right)^{2}=y(x)\cdot y(x)$ (the product rule has to be applied here a second time)
• $f'(x)=1$
• $g'(x)=y'(x)\cdot y(x)+y(x)\cdot y'(x)$
• $\left(f(x)\cdot g(x)\right)^{'}=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
• $\left(f(x)\cdot g(x)\right)^{'}=1\cdot\left(y(x)\right)^{2}+x\cdot\left(y'(x)\cdot y(x)+y(x)\cdot y'(x)\right)$
• $\left(f(x)\cdot g(x)\right)^{'}=\left(y(x)\right)^{2}+2\cdot x\cdot y'(x)\cdot y(x)$