Given the Differential equation $y'=-2xy^{2}$.
Find the derivative $\frac{d(y')}{dx}$!
My approach, which is not correct according to Wolfram Alpha:
- Plugging in: $\frac{d(y')}{dx}=\frac{d(-2xy^{2})}{dx}$
- Pulling out constants #1:$-2\cdot\frac{d(x^{1}\cdot y^{2})}{dx}$
- Pulling out constants #2:$-2\cdot y^{2}\cdot\frac{d(x^{1})}{dx}$
- 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)
- $f'(x)=1$
- $g'(x)=y'(x)\cdot y(x)+y(x)\cdot y'(x)$
- $\left(f(x)\cdot g(x)\right)^{'}=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
- $\left(f(x)\cdot g(x)\right)^{'}=1\cdot\left(y(x)\right)^{2}+x\cdot\left(y'(x)\cdot y(x)+y(x)\cdot y'(x)\right)$
- $\left(f(x)\cdot g(x)\right)^{'}=\left(y(x)\right)^{2}+2\cdot x\cdot y'(x)\cdot y(x)$
2 Answers
$\begingroup$Keep in mind that $y$ is not a constant, it is a function of $x$. Therefore you have to use the product rule between $x$ and $y^2$.
$\endgroup$ 3 $\begingroup$Pulling out constants #2:$-2\cdot y^{2}\cdot\frac{d(x^{1})}{dx}$
Nope that's wrong, you had in step 2, $$-2\frac{d(x^{1}\cdot y^{2})}{dx}$$ which can be expanded by product rule as $$-2y^{2}\frac{d(x^{1})}{dx}-2x^1\frac{d(y^{2})}{dx}$$ Now that's, $$-2y^{2}1x^0-2x^1\frac{d(y^{2})}{dy}.\frac{y}{x}=2y^2-2x.2y.y'=2y^2-4xy(-2xy^2)=2y^2+8x^2y^3$$ by chain rule and substitution. The reason being y is a dependent variable, yes a variable.
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