Question

Обчисліть значення похідної даної функції в точці х0 1)f(x)=x^2-4x+1,x0=1 2)f(x)cosx+sinx,x0 = π/4 3) f(x)=2x+√x, x0=9

Answer 1

Ответ:

1

$f(x) = {x}^{2} - 4x + 1$

$f'(x) = 2x - 4$

$f'(1) = 2 - 4 = - 2$

2

$f(x) = \cos(x) + \sin(x)$

$f'(x) = - \sin(x) + \cos(x)$

$f'( \frac{\pi}{4} ) = - \sin( \frac{\pi}{4} ) + \cos( \frac{\pi}{4} ) = - \frac{{ \sqrt{2} } }{2} + \frac{ \sqrt{2} }{2} = 0$

3

$f(x) = 2x + \sqrt{x} = 2x + {x}^{ \frac{1}{2} }$

$f'(x) = 2 + \frac{1}{2} {x}^{ - \frac{1}{2} } = 2 + \frac{1}{2 \sqrt{x} }$

$f'(9) = 2 + \frac{1}{2 \times 3} = 2 + \frac{1}{6} = 2 \frac{1}{6}$

Answer 2

1) f(x) = x^2 - 4x + 1, x0 = 1

To find the derivative of the given function f(x) = x^2 - 4x + 1 at the point x0 = 1, we can use the power rule of differentiation. The power rule states that if we have a function of the form f(x) = x^n, then its derivative is given by f'(x) = nx^(n-1).

Here, the function f(x) = x^2 - 4x + 1, so we can differentiate each term separately:

f'(x) = (2x^1) - (4x^0) + (0) = 2x - 4

Now, to find the value of the derivative at x0 = 1, we substitute x = 1 into the derivative expression:

f'(1) = 2(1) - 4 = -2

Therefore, the value of the derivative of f(x) = x^2 - 4x + 1 at x0 = 1 is -2.

2) f(x) = cos(x) + sin(x), x0 = π/4

To find the derivative of the given function f(x) = cos(x) + sin(x) at the point x0 = π/4, we can again use the power rule and the derivative rules for trigonometric functions.

The derivative of cos(x) is -sin(x), and the derivative of sin(x) is cos(x). Therefore, the derivative of f(x) = cos(x) + sin(x) is:

f'(x) = -sin(x) + cos(x)

Now, let's find the value of the derivative at x0 = π/4:

f'(π/4) = -sin(π/4) + cos(π/4)

Using the values of sin(π/4) = sqrt(2)/2 and cos(π/4) = sqrt(2)/2, we can simplify the expression:

f'(π/4) = -(sqrt(2)/2) + sqrt(2)/2

The two terms cancel out, and we're left with:

f'(π/4) = 0

Therefore, the value of the derivative of f(x) = cos(x) + sin(x) at x0 = π/4 is 0.

3) f(x) = 2x + √x, x0 = 9

To find the derivative of the given function f(x) = 2x + √x at the point x0 = 9, we can again apply the power rule and the chain rule.

The derivative of 2x is simply 2. For the term √x, we can rewrite it as x^(1/2) and apply the power rule to get (1/2)x^(-1/2).

Therefore, the derivative of f(x) = 2x + √x is:

f'(x) = 2 + (1/2)x^(-1/2)

Now, let's find the value of the derivative at x0 = 9:

f'(9) = 2 + (1/2)(9)^(-1/2)

To simplify the expression, we need to calculate (9)^(-1