Dy y dx

Dy y dx

Try using the chain rule for exponents. That’s

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

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

[math]u=y^2\implies[/math]

[math]\dfrac{du}{dx}=\dfrac{d}{dx}y^2=\dfrac{d}{dx}y(x)^2[/math]

[math]=2y(x)^{2-1}\dfrac{d}{dx}y(x)[/math]

[math]=2y(x)^1\dfrac{d}{dx}y(x)[/math]

[math]=2y(x)\dfrac{d}{dx}y(x)[/math]

Now remove the [math]x[/math]s and get

[math]\dfrac{dy}{dx}=2y\dfrac{dy}{dx}[/math]



Источник: www.quora.com


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