Question

Очень срочно!!! Найти тангенс угла наклона касательной к графику функции y = sinx в точке с абсциссой х0=П/2пожалуйста только с объяснениями.Заранее спасибо

Answer 1

тангенс угла наклона касательной к графику функции равен производной функции в точке х нулевое.
1)Находим производную функции:
   y=sinx
   y`=cosx
2)Находим значение производной в точке х0:
  y`(x0)=y`(п/2)=cos(п/2)=0
3) $tg \alpha=0$

Answer 2

Finding the Tangent of the Slope of a Function

To find the tangent of the slope of a function at a given point, we need to calculate the derivative of the function and evaluate it at that point. In this case, we want to find the tangent of the slope of the function y = sin(x) at the point with x-coordinate x0 = π/2.

The derivative of the function y = sin(x) can be found using the chain rule. The derivative of sin(x) is cos(x). Therefore, the derivative of y = sin(x) is dy/dx = cos(x).

To find the tangent of the slope at x0 = π/2, we need to evaluate the derivative dy/dx = cos(x) at x = π/2.

Let's calculate it step by step:

1. Find the derivative of y = sin(x): - The derivative of sin(x) is cos(x).

2. Evaluate the derivative at x = π/2: - Substitute x = π/2 into the derivative dy/dx = cos(x): dy/dx = cos(π/2) = 0.

Therefore, the tangent of the slope of the function y = sin(x) at the point with x-coordinate x0 = π/2 is 0.

Please note that the tangent of the slope represents the rate of change of the function at that point. In this case, the function y = sin(x) has a horizontal tangent at x = π/2, indicating that the rate of change of the function is 0 at that point.

I hope this explanation helps! Let me know if you have any further questions.