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The derivative of the upper limit $u=\sqrt{t}$ with respect to $t$ is $1/(2\sqrt{t})$. This factor is used in the chain rule when differentiating an integral with this variable upper limit.

Conditions: $t > 0$.; The upper limit is defined as $u=\sqrt{t}$.

The derivative of the upper limit $u=\sqrt{t}$ with respect to $t$ is $1/(2\sqrt{t})$. This factor is used in the chain rule when differentiating an integral with this variable upper limit.

Conditions: $t > 0$.; The upper limit is defined as $u=\sqrt{t}$.

To differentiate an integral with a square-root upper limit, you evaluate the integrand at the upper limit and multiply by the derivative of that limit. For an upper limit $u=\sqrt{t}$, the derivative is $1/(2\sqrt{t})$.

Conditions: The integrand is continuous on the relevant interval.; The upper limit is a differentiable function of $t$, specifically $u=\sqrt{t}$.; $t > 0$.

To differentiate an integral with a square-root upper limit, you evaluate the integrand at the upper limit and multiply by the derivative of that limit. For an upper limit $u=\sqrt{t}$, the derivative is $1/(2\sqrt{t})$.

Conditions: The integrand is continuous on the relevant interval.; The upper limit is a differentiable function of $t$, specifically $u=\sqrt{t}$.; $t > 0$.

The factor is essential because the Fundamental Theorem of Calculus combined with the Chain Rule requires multiplying the integrand evaluated at the upper limit by the derivative of that limit. Without this factor, the rate of change of the accumulated area with respect to the outer variable $t$ would be incorrect.

Conditions: The upper limit is a function of $t$, not just $t$ itself.; The integrand is continuous.

The factor is essential because the Fundamental Theorem of Calculus combined with the Chain Rule requires multiplying the integrand evaluated at the upper limit by the derivative of that limit. Without this factor, the rate of change of the accumulated area with respect to the outer variable $t$ would be incorrect.

Conditions: The upper limit is a function of $t$, not just $t$ itself.; The integrand is continuous.