F(x)=int_(0)^(sqrtx)2tdt where x > 0. F^(')(x)=

Apply Fundamental Theorem: Use the Fundamental Theorem of Calculus Part 1 to find F'(x). This theorem states that if F(x) is defined as an integral from a to g(x) of f(t) dt, then F'(x) is f(g(x)) times g'(x).Find Derivative of sqrt(x): First, find the derivative of the upper limit of integration, which is sqrt(x). The derivative of sqrt(x) with respect to x is (1/2)x^(-1/2).Evaluate Integrand at sqrt(x): Now, apply the Fundamental Theorem of Calculus. The integrand is 2t, and we need to evaluate it at the upper limit, which is sqrt(x). So we replace t with sqrt(x) to get 2sqrt(x).Multiply Result by Derivative: Multiply the result from the previous step by the derivative of the upper limit. So we have 2sqrt(x) * (1/2)x^(-1/2).Simplify the Expression: Simplify the expression. The (1/2)x^(-1/2) cancels out the sqrt(x) in 2sqrt(x), leaving us with just 2.

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F(x)=int_(0)^(sqrtx)2tdt
where
x > 0.

F^(')(x)=

$F(x)=\int_{0}^{\sqrt{x}} 2 t d t$ $\newline$ where x>0 . $\newline$ $F^{\prime}(x)=$

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Evaluate definite integrals using the power rule

Full solution

F(x)=\int_{0}^{\sqrt{x}} 2 t d t

\newline

where

x>0

\newline

F^{\prime}(x)=

Apply Fundamental Theorem: Use the Fundamental Theorem of Calculus Part $1$ to find $F'(x)$ . This theorem states that if $F(x)$ is defined as an integral from $a$ to $g(x)$ of $f(t) \, dt$ , then $F'(x)$ is $f(g(x)) \times g'(x)$ .
Find Derivative of $\sqrt{x}$ : First, find the derivative of the upper limit of integration, which is $\sqrt{x}$ . The derivative of $\sqrt{x}$ with respect to $x$ is $(1/2)x^{(-1/2)}$ .
Evaluate Integrand at $\sqrt{x}$ : Now, apply the Fundamental Theorem of Calculus. The integrand is $2t$ , and we need to evaluate it at the upper limit, which is $\sqrt{x}$ . So we replace $t$ with $\sqrt{x}$ to get $2\sqrt{x}$ .
Multiply Result by Derivative: Multiply the result from the previous step by the derivative of the upper limit. So we have $2\sqrt{x} \times \left(\frac{1}{2}x^{-\frac{1}{2}}\right)$ .
Simplify the Expression: Simplify the expression. The $(\frac{1}{2})x^{(-\frac{1}{2})}$ cancels out the $\sqrt{x}$ in $2\sqrt{x}$ , leaving us with just $2$ .