Respuesta :

caylus

Answer:

Hello,

[tex]\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]

Step-by-step explanation:

[tex](f(x)+g(x))'=f'(x)+g'(x)\\\\(2x)'=2*(x)'=2*1=2\\\\(\dfrac{200}{x} )'=200*(x^{-1})'=200*(-1)*x^{-1-1})=-\dfrac{200}{x^2} \\\\\\\boxed{y'=2-\dfrac{200}{x^2} }\\[/tex]

etanauriel

Answer:

[tex] \frac{dy}{dx } = 2 - \frac{200}{ {x}^{2} } [/tex]

Step-by-step explanation:

[tex]the \: equation \: can \: be \: rewriten \: as \\ y = 2x + 200 {x}^{ - 1} \\ \\ now \: differentiate \: the \: equation\ \\ \frac{dy}{dx} = 2 - 200 {x}^{ - 2} \\ \frac{dy}{dx} = 2 - \frac{200}{ {x}^{2} } [/tex]

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