Let g:[c, d] →ℝ be such that g([c, d]) ⊆[a, b], and let f:[a, b] →ℝ be integrable. Define G:[c,

step 1 So they said let F be differentiable on the open interval AB. F of A and F of B equal to zero. A

Let g:[c, d] →ℝ be such that g([c, d]) ⊆[a, b], and let...

Let $g:[c, d] \rightarrow \mathbb{R}$ be such that $g([c, d]) \subseteq[a, b]$, and let $f:[a, b] \rightarrow \mathbb{R}$ be integrable. Define $G:[c, d] \rightarrow \mathbb{R}$ by $$ G(y):=\int_{a}^{g(y)} f(t) d t $$ If $g$ is differentiable at $y_{0} \in[c, d]$ and $f$ is continuous at $g\left(y_{0}\right)$, then show that $G$ is differentiable at $y_{0}$ and $G^{\prime}\left(y_{0}\right)=f\left(g\left(y_{0}\right)\right) g^{\prime}\left(y_{0}\right)$.

Let $g:[c, d] \rightarrow \mathbb{R}$ be such that $g([c, d]) \subseteq[a, b]$, and let $f:[a, b] \rightarrow \mathbb{R}$ be
integrable. Define $G:[c, d] \rightarrow \mathbb{R}$ by
$$
G(y):=\int_{a}^{g(y)} f(t) d t
$$
If $g$ is differentiable at $y_{0} \in[c, d]$ and $f$ is continuous at $g\left(y_{0}\right)$, then show that $G$ is differentiable at $y_{0}$ and $G^{\prime}\left(y_{0}\right)=f\left(g\left(y_{0}\right)\right) g^{\prime}\left(y_{0}\right)$.

Key Concepts

Differentiation of Functions Defined by Integrals

When a function is defined as an integral with a variable limit, its differentiability can often be established under suitable conditions such as continuity of the integrand and differentiability of the limit function. This process leverages both the fundamental theorem of calculus and the chain rule to compute the derivative at a point.

Continuity Conditions

Continuity of the integrand is a key condition required to apply the fundamental theorem of calculus. Ensuring that the integrand is continuous at the point corresponding to the variable limit maintains the integrity of the differentiation process of the integral, guaranteeing that small changes in the variable lead to small, controlled changes in the integral’s value.

Chain Rule

The chain rule is a fundamental tool in calculus used to compute the derivative of a composite function. In the context of an integral with a variable limit that itself is a function, the derivative is obtained by multiplying the derivative of the outer function (given by the fundamental theorem of calculus) by the derivative of the inner function (the variable limit).

Fundamental Theorem of Calculus

This theorem connects differentiation and integration by stating that if a function is continuous over an interval, then the function defined by its integral with a variable upper limit is differentiable, and its derivative is the original function evaluated at that upper limit. This is a crucial concept in the analysis of functions defined via integrals.

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