Question

Y=(3x)^-1/2+(3x)^1/2 найти y'(1/3)

Answer 1

Ответ: Y=(3x)^-1/2+(3x)^1/2 найти y'(1/3)

Пошаговое объяснение:

Answer 2

To find $y'(1/3)$, we need to differentiate the given function $y = (3x)^{-1/2} + (3x)^{1/2}$ with respect to $x$. Let's proceed with the differentiation:

First, we'll differentiate the first term $(3x)^{-1/2}$ using the power rule for differentiation. According to the power rule, if $f(x) = ax^n$, then $f'(x) = anx^{n-1}$. Applying this rule:

$\frac{d}{dx}[(3x)^{-1/2}] = -\frac{1}{2}(3x)^{-1/2-1} \cdot \frac{d}{dx}(3x) = -\frac{1}{2}(3x)^{-3/2} \cdot 3$

Simplifying this, we get:

$\frac{d}{dx}[(3x)^{-1/2}] = -\frac{3}{2}(3x)^{-3/2}$

Now, let's differentiate the second term $(3x)^{1/2}$ using the power rule:

$\frac{d}{dx}[(3x)^{1/2}] = \frac{1}{2}(3x)^{1/2-1} \cdot \frac{d}{dx}(3x) = \frac{1}{2}(3x)^{-1/2} \cdot 3$

Simplifying this, we have:

$\frac{d}{dx}[(3x)^{1/2}] = \frac{3}{2}(3x)^{-1/2}$

Now, to find $y'(1/3)$, we substitute $x = 1/3$ into the derivatives we found above:

$y'(1/3) = -\frac{3}{2}(3(1/3))^{-3/2} + \frac{3}{2}(3(1/3))^{-1/2}$

Simplifying further:

$y'(1/3) = -\frac{3}{2}(1)^{-3/2} + \frac{3}{2}(1)^{-1/2}$

Since $1^n = 1$ for any $n$, we have:

$y'(1/3) = -\frac{3}{2}(1) + \frac{3}{2}(1) = 0$

Therefore, the value of $y'(1/3)$ is 0.