Dy y dx

# Dy y dx

Try using the chain rule for exponents. That’s

$\dfrac{d}{dx}u^n=nu^{n-1}\dfrac{du}{dx}\text{,}\space n\neq 0\text{.}$

You have to assume that $y$ is a function of $x$, that is $y=y(x)$ because you are differentiating w.r.t. $x$ and not $y$.

$u=y^2\implies$

$\dfrac{du}{dx}=\dfrac{d}{dx}y^2=\dfrac{d}{dx}y(x)^2$

$=2y(x)^{2-1}\dfrac{d}{dx}y(x)$

$=2y(x)^1\dfrac{d}{dx}y(x)$

$=2y(x)\dfrac{d}{dx}y(x)$

Now remove the $x$s and get

$\dfrac{dy}{dx}=2y\dfrac{dy}{dx}$

Источник: www.quora.com

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