Question

Y=x+4/x дифференцирования​

Answer 1

Ответ:

Объяснение:

$y(x)=\frac{x+4}{x}$

$y'(x)=(\frac{x}{x} +\frac{4}{x} )'=(1)'+\frac{4}{x} '=-\frac{4}{x^2}$

Answer 2

To differentiate the function $y = \frac{x + 4}{x}$ with respect to $x$, we can use the quotient rule. The quotient rule states that if you have a function of the form $y = \frac{u}{v}$, where $u$ and $v$ are both functions of $x$, then the derivative $\frac{dy}{dx}$ is given by:

$\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

In this case, $u = x + 4$ and $v = x$. Let's differentiate $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = 1$ $\frac{dv}{dx} = 1$

Now we can plug these values into the quotient rule formula:

$\frac{dy}{dx} = \frac{x \cdot 1 - (x + 4) \cdot 1}{x^2}$ $\frac{dy}{dx} = \frac{x - x - 4}{x^2}$ $\frac{dy}{dx} = \frac{-4}{x^2}$

So, the derivative of $y = \frac{x + 4}{x}$ with respect to $x$ is $\frac{-4}{x^2}$.