Why is the reciprocal endpoint-derivative factor essential when differentiating an integral with a variable upper limit?

The model tracks a small input change $dx$ through successive local linear approximations. First, the inner function $h(x)=x^2$ scales the input change by its derivative $h'(x)=2x$.

Conditions: Functions $h$ and $g$ are differentiable at the relevant points.; The input change $dx$ is infinitesimal.; The outer derivative is evaluated at the inner function's output $h(x)$, not at $x$.

The chain rule holds whenever both the inner function $h$ is differentiable at $x$ and the outer function $g$ is differentiable at $h(x)$. If $h'(x)=0$, the formula $g'(h(x))h'(x)$ correctly yields 0.

Conditions: $h$ is differentiable at $x$; $g$ is differentiable at $h(x)$; No requirement for $h'(x) \neq 0$

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 $u=\sqrt{t}$.

By tracking a small input change $dx$ through nested mappings $x \to h(x) \to g(h(x))$. For $x=1.5$, $h=x^2$, and $g=\sin$, the inner rate is $h'(1.5)=3$, scaling $dx$ to $dh \approx 3dx$.

Conditions: Inner function $h(x)$ is differentiable at $x$; Outer function $g(u)$ is differentiable at $u=h(x)$; Example uses $h(x)=x^2$ and $g(u)=\sin u$ near $x=1.5$

The simplified derivative is $\sin(t^{5/2})/(2t^{5/2})$. This is obtained by evaluating the integrand at $x=\sqrt{t}$ and multiplying by the derivative of the upper limit, $1/(2\sqrt{t})$.

Conditions: The integrand is $\sin(x^5)/x^4$.; The lower limit is the constant 2.; The upper limit is $\sqrt{t}$.; $t > 0$.