Question

Производная для у = х - 1 / х + 4

Answer 1

Производную заказывали? Пожалуйста: y'=(x-x^(-1)+4)'=1+x^(-2)

Ответ: y=1+1/x^2.

Удачи!

Answer 2

$y = x - \dfrac{1}{x} + 4$

$y' = 1 - \bigg(-\dfrac{1}{x^{2}} \bigg) = \dfrac{x^{2} + 1}{x^{2}}$

Answer 3

To find the derivative of the given function $y = \frac{x - 1}{x + 4}$, we can use the quotient rule of differentiation. The quotient rule states that if we have a function $y = \frac{u(x)}{v(x)}$, then its derivative is given by:

$\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$

where $u'(x)$ and $v'(x)$ are the derivatives of $u(x)$ and $v(x)$, respectively.

Let's find the derivative of $y$ step by step:

1. Identify $u(x)$ and $v(x)$ in the given function: $u(x) = x - 1$ $v(x) = x + 4$

2. Find $u'(x)$ and $v'(x)$: The derivative of $u(x) = x - 1$ with respect to $x$ is $u'(x) = 1$, since the derivative of $x$ with respect to itself is 1, and the derivative of a constant (in this case, -1) is zero.

   The derivative of $v(x) = x + 4$ with respect to $x$ is $v'(x) = 1$, since the derivative of $x$ with respect to itself is 1, and the derivative of a constant (in this case, 4) is zero.

3. Apply the quotient rule to find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{(x + 4) \cdot 1 - (x - 1) \cdot 1}{(x + 4)^2}$ $\frac{dy}{dx} = \frac{x + 4 - x + 1}{(x + 4)^2}$ $\frac{dy}{dx} = \frac{5}{(x + 4)^2}$

So, the derivative of $y = \frac{x - 1}{x + 4}$ with respect to $x$ is $\frac{5}{(x + 4)^2}$.