What is the value of (d)/(dx)(root(5)(x^(2))) at x=32 ?

Rewrite function: We need to find the derivative of the function f(x) = x^(2/5). To do this, we will use the power rule for differentiation, which states that if f(x) = x^n, then f'(x) = n*x^(n-1).Apply power rule: First, we rewrite the fifth root of x squared as x^(2/5) to make it easier to differentiate. f(x) = (x^2)^(1/5) = x^(2/5)Find derivative: Now we apply the power rule for differentiation to find f'(x). f'(x) = (2/5)*x^((2/5)-1) = (2/5)*x^(-3/5)Evaluate at x=32: Next, we need to evaluate the derivative at x=32. f'(32) = (2/5)*32^(-3/5)Simplify expression: We simplify the expression by calculating 32^(-3/5). Since 32 is 2^5, we can rewrite 32^(-3/5) as (2^5)^(-3/5). 32^(-3/5) = (2^5)^(-3/5) = 2^(5*(-3/5)) = 2^(-3)Calculate value: Now we calculate 2^(-3), which is 1/(2^3) = 1/8. 2^(-3) = 1/(2^3) = 1/8Finally, we multiply (2/5) by (1/8) to get the value of the derivative at x=32. f'(32) = (2/5)*(1/8) = 2/(5*8) = 2/40 = 1/20

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What is the value of
(d)/(dx)(root(5)(x^(2))) at
x=32 ?

What is the value of $\frac{d}{d x}\left(\sqrt[5]{x^{2}}\right)$ at $x=32$ ?

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Algebra 1

Multiplication with rational exponents

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Q. What is the value of

\frac{d}{d x}\left(\sqrt[5]{x^{2}}\right)

x=32

Rewrite function: We need to find the derivative of the function $f(x) = x^{\frac{2}{5}}$ . To do this, we will use the power rule for differentiation, which states that if $f(x) = x^n$ , then $f'(x) = n \cdot x^{n-1}$ .
Apply power rule: First, we rewrite the fifth root of $x$ squared as $x^{(2/5)}$ to make it easier to differentiate. $\newline$ $f(x) = (x^2)^{(1/5)} = x^{(2/5)}$
Find derivative: Now we apply the power rule for differentiation to find $f'(x)$ . $f'(x) = \left(\frac{2}{5}\right)*x^{\left(\frac{2}{5}-1\right)} = \left(\frac{2}{5}\right)*x^{-\frac{3}{5}}$
Evaluate at $x=32$ : Next, we need to evaluate the derivative at $x=32$ . $f'(32) = \left(\frac{2}{5}\right)\cdot32^{-\frac{3}{5}}$
Simplify expression: We simplify the expression by calculating $32^{(-3/5)}$ . Since $32$ is $2^5$ , we can rewrite $32^{(-3/5)}$ as $(2^5)^{(-3/5)}$ . $\newline$ $32^{(-3/5)} = (2^5)^{(-3/5)} = 2^{5*(-3/5)} = 2^{-3}$
Calculate value: Now we calculate $2^{-3}$ , which is $1/(2^3) = 1/8$ . $\newline$ $2^{-3} = 1/(2^3) = 1/8$
Calculate value: Now we calculate $2^{-3}$ , which is $1/(2^3) = 1/8$ . $2^{-3} = 1/(2^3) = 1/8$ Finally, we multiply $(2/5)$ by $(1/8)$ to get the value of the derivative at $x=32$ . $f'(32) = (2/5)*(1/8) = 2/(5*8) = 2/40 = 1/20$