Question

F (x)=(x^2-2)корень из х^2+1 найти f'(корень из 3) пожалуйста

Answer 1

Решение в прикреплённом файле

Answer 2

Calculating the Derivative of \( f(x) = \sqrt{x^2 + 1} \)

To find the derivative of \( f(x) = \sqrt{x^2 + 1} \) at \( x = \sqrt{3} \), we can use the chain rule. The chain rule states that if \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \cdot h'(x) \).

The derivative of \( f(x) = \sqrt{x^2 + 1} \) can be found using the chain rule as follows:

1. Let \( u = x^2 + 1 \). 2. Then \( f(x) = \sqrt{u} \). 3. Using the chain rule, the derivative of \( f(x) \) with respect to \( x \) is given by: \[ f'(x) = \frac{1}{2\sqrt{u}} \cdot \frac{d(u)}{dx} \]

Now, let's find the derivative of \( u = x^2 + 1 \): \[ \frac{d(u)}{dx} = 2x \]

Substitute \( u = x^2 + 1 \) and \( \frac{d(u)}{dx} = 2x \) back into the expression for \( f'(x) \): \[ f'(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x \]

Finally, we can evaluate \( f'(x) \) at \( x = \sqrt{3} \): \[ f'(\sqrt{3}) = \frac{1}{2\sqrt{(\sqrt{3})^2 + 1}} \cdot 2\sqrt{3} \]

Result:

The derivative of \( f(x) = \sqrt{x^2 + 1} \) at \( x = \sqrt{3} \) is: \[ f'(\sqrt{3}) = \frac{\sqrt{3}}{2\sqrt{4}} = \frac{\sqrt{3}}{4} \]

Therefore, the value of \( f'(\sqrt{3}) \) is \( \frac{\sqrt{3}}{4} \).