Question

F(x)=корень(5-х) f''(1)

Answer 1

Ответ:

$f(x)=\sqrt{5-x}\\\\f'(x)=\dfrac{-1}{2\sqrt{5-x}}\\\\\\f''(x)=-\dfrac{1}{2}\cdot \Big(-\dfrac{1}{2}\Big)\cdot (5-x)^{-\frac{3}{2}}\cdot (-1)=\dfrac{-1}{4\sqrt{(5-x)^3}}\\\\\\f''(1)=-\dfrac{1}{4\cdot 8}=-\dfrac{1}{32}$

Answer 2

To find the second derivative, $f''(x)$, of the function $f(x) = \sqrt{5 - x}$, you'll need to first find the first derivative, $f'(x)$, and then take the derivative of $f'(x)$ to find $f''(x)$.

Let's start with the first derivative:

$f(x) = \sqrt{5 - x}$

Using the chain rule, we can find $f'(x)$:

$f'(x) = \frac{d}{dx}(\sqrt{5 - x})$

Let $u = 5 - x$, so $f'(x) = \frac{d}{du}(\sqrt{u}) \cdot \frac{du}{dx}$

Now, find the derivative of $\sqrt{u}$ with respect to $u$:

$\frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}}$

And find $\frac{du}{dx}$ by taking the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = -1$ (since the derivative of $-x$ is $-1$)

Now, combine these results:

$f'(x) = \frac{1}{2\sqrt{u}} \cdot (-1) = -\frac{1}{2\sqrt{u}}$

Now, you need to evaluate this derivative at $x = 1$ to find $f''(1)$:

$f''(1) = -\frac{1}{2\sqrt{5 - 1}} = -\frac{1}{2\sqrt{4}} = -\frac{1}{2 \cdot 2} = -\frac{1}{4}$

So, $f''(1) = -\frac{1}{4}$.