Question

Производная функции: 2/5 корень из X

Answer 1

Y = 2/5*√х;
у ' = 2/5*(√х)' = 2/5*1/2√х = 2/10*√х = 0,2√х.

1.(√х)' = 1/2√х.
2.постоянный множитель 2/5 можно вынести за знак производной.

Answer 2

To find the derivative of the given function $f(x) = \frac{2}{5} \sqrt{x}$ with respect to $x$, we can use the rules of differentiation. The derivative of a function represents its rate of change or the slope of the tangent line at any given point.

To differentiate $f(x) = \frac{2}{5} \sqrt{x}$ with respect to $x$, we'll use the power rule and the chain rule.

The power rule states that if we have a function of the form $g(x) = x^n$, then its derivative with respect to $x$ is $g'(x) = n \cdot x^{n-1}$.

The chain rule states that if we have a function of the form $h(x) = f(g(x))$, then its derivative with respect to $x$ is $h'(x) = f'(g(x)) \cdot g'(x)$.

Now, let's apply these rules to find the derivative of $f(x) = \frac{2}{5} \sqrt{x}$:

Let $g(x) = x$ (inside the square root) and $f(g) = \sqrt{g} = g^{1/2}$.

1. Find $g'(x)$: $g'(x) = 1$, since the derivative of $x$ with respect to $x$ is 1.

2. Find $f'(g)$: $f'(g) = \frac{d}{dg} (g^{1/2}) = \frac{1}{2} g^{-1/2}$ (using the power rule).

3. Combine the results using the chain rule: $f'(x) = f'(g) \cdot g'(x) = \frac{1}{2} \cdot x^{-1/2} = \frac{1}{2\sqrt{x}}$.

So, the derivative of the function $f(x) = \frac{2}{5} \sqrt{x}$ with respect to $x$ is $f'(x) = \frac{1}{2\sqrt{x}}$.