What is the value of (d)/(dx)(sqrt(x^(3))) at x=25 ?

Rewrite function: We need to find the derivative of the function f(x) = sqrt(x^3) with respect to x. The function can be rewritten as f(x) = (x^3)^(1/2).Apply chain rule: To differentiate f(x) = (x^3)^(1/2), we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Find outer function derivative: The outer function is g(u) = u^(1/2) and the inner function is u(x) = x^3. We will first find the derivative of the outer function, g'(u) = (1/2)u^(-1/2).Find inner function derivative: Next, we find the derivative of the inner function, u'(x) = 3x^2.Apply chain rule again: Now we apply the chain rule: f'(x) = g'(u(x)) * u'(x). Substituting the derivatives we found, we get f'(x) = (1/2)(x^3)^(-1/2) * 3x^2.Simplify expression: Simplify the expression: f'(x) = (3/2)x^2 * (x^3)^(-1/2).Rewrite with single exponent: Rewrite the expression with a single exponent: f'(x) = (3/2)x^(2 - 3/2) = (3/2)x^(1/2).Evaluate derivative at x=25: Now we evaluate the derivative at x = 25: f'(25) = (3/2)(25)^(1/2).Calculate final value: Since the square root of 25 is 5, we have f'(25) = (3/2) * 5.Finally, we calculate the value: f'(25) = (3/2) * 5 = 15/2 = 7.5.

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What is the value of
(d)/(dx)(sqrt(x^(3))) at
x=25 ?

What is the value of $\frac{d}{d x}\left(\sqrt{x^{3}}\right)$ at $x=25$ ?

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Algebra 1

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Q. What is the value of

\frac{d}{d x}\left(\sqrt{x^{3}}\right)

x=25

Rewrite function: We need to find the derivative of the function $f(x) = \sqrt{x^3}$ with respect to $x$ . The function can be rewritten as $f(x) = (x^3)^{\frac{1}{2}}$ .
Apply chain rule: To differentiate $f(x) = (x^3)^{\frac{1}{2}}$ , we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Find outer function derivative: The outer function is $g(u) = u^{\frac{1}{2}}$ and the inner function is $u(x) = x^3$ . We will first find the derivative of the outer function, $g'(u) = \left(\frac{1}{2}\right)u^{-\frac{1}{2}}$ .
Find inner function derivative: Next, we find the derivative of the inner function, $u'(x) = 3x^2$ .
Apply chain rule again: Now we apply the chain rule: $f'(x) = g'(u(x)) \cdot u'(x)$ . Substituting the derivatives we found, we get $f'(x) = \left(\frac{1}{2}\right)\left(x^3\right)^{-\frac{1}{2}} \cdot 3x^2$ .
Simplify expression: Simplify the expression: $f'(x) = \left(\frac{3}{2}\right)x^2 \cdot \left(x^3\right)^{-\frac{1}{2}}$ .
Rewrite with single exponent: Rewrite the expression with a single exponent: $f'(x) = \left(\frac{3}{2}\right)x^{2 - \frac{3}{2}} = \left(\frac{3}{2}\right)x^{\frac{1}{2}}$ .
Evaluate derivative at $x=25$ : Now we evaluate the derivative at $x = 25$ : $f'(25) = \left(\frac{3}{2}\right)(25)^{\frac{1}{2}}$ .
Calculate final value: Since the square root of $25$ is $5$ , we have $f'(25) = (\frac{3}{2}) \times 5$ .
Calculate final value: Since the square root of $25$ is $5$ , we have $f'(25) = (\frac{3}{2}) \times 5$ .Finally, we calculate the value: $f'(25) = (\frac{3}{2}) \times 5 = \frac{15}{2} = 7.5$ .