Question

Укажите значение производной функции y=e^xcosx в точке x0=0

Answer 1

$f ' =(e^{x} *cosx) ' =( e^{x} ) ' *cosx+ e^{x} *(cosx) ' =$
$e^{x} cosx+e^x( - sinx)= e^x*(cosx-sinx)$
$= e^0*(cos0-sin0)=1*(1-0)=1$

Answer 2

Finding the Derivative of the Function

To find the derivative of the function \( y = e^x \cdot \cos(x) \) at the point \( x_0 = 0 \), we can use the product rule of differentiation, which states that if \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative of their product is given by:

\[ (u \cdot v)' = u'v + uv' \]

Applying the Product Rule

Let's apply the product rule to find the derivative of the given function \( y = e^x \cdot \cos(x) \) at the point \( x_0 = 0 \):

1. First, we find the derivative of \( e^x \), denoted as \( u(x) \), which is \( e^x \). 2. Then, we find the derivative of \( \cos(x) \), denoted as \( v(x) \), which is \( -\sin(x) \).

Now, we can apply the product rule:

\[ y' = (e^x \cdot \cos(x))' = e^x \cdot (-\sin(x)) + e^x \cdot \cos(x) \]

Evaluating at \( x_0 = 0 \)

Now, we can evaluate the derivative at the point \( x_0 = 0 \):

\[ y'(0) = e^0 \cdot (-\sin(0)) + e^0 \cdot \cos(0) = 1 \cdot 0 + 1 \cdot 1 = 1 \]

Therefore, the value of the derivative of the function \( y = e^x \cdot \cos(x) \) at the point \( x_0 = 0 \) is 1.