Question

Найдите скорость изменения функции y=2cos(2x-2) в точке x0=1

Answer 1

Скорость изменения функции - это производная от неё

y ' = 2 (cos (2x - 2)) ' = - 2 sin (2x - 2) (2x - 2) ' = - 4 sin (2x - 2)

y ' (1) = - 4 sin (2*1 - 2) = - 4 sin 0 = 0 

Answer 2

Finding the Derivative of the Function

To find the derivative of the function \( y = 2\cos(2x-2) \) at the point \( x_0 = 1 \), we can use the chain rule for differentiation. The chain rule states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where \( u = g(x) \).

Applying the Chain Rule

In this case, let's consider \( u = 2x-2 \) and \( y = 2\cos(u) \). The derivative of \( y \) with respect to \( u \) is \( \frac{dy}{du} = -2\sin(u) \), and the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = 2 \).

Calculating the Derivative at \( x_0 = 1 \)

Now, we can calculate the derivative of the function \( y = 2\cos(2x-2) \) at the point \( x_0 = 1 \) by substituting \( x_0 = 1 \) into the derivatives we found.

The derivative of the function at \( x_0 = 1 \) is: \[ \frac{dy}{dx} \bigg|_{x=1} = \frac{dy}{du} \cdot \frac{du}{dx} \bigg|_{x=1} = -2\sin(u) \cdot 2 \bigg|_{x=1} \]

Result

Therefore, the derivative of the function \( y = 2\cos(2x-2) \) at the point \( x_0 = 1 \) is: \[ \frac{dy}{dx} \bigg|_{x=1} = -4\sin(0) = 0 \]

The derivative of the function \( y = 2\cos(2x-2) \) at the point \( x_0 = 1 \) is 0. This means that at \( x = 1 \), the function has a horizontal tangent and is not changing with respect to \( x \) at that point.

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