Q. 17 - Calculus

Question

Explain precisely how the formula $\frac{d}{dx}{\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\text{'}}\left(x\right)$in Theorem $4.35$ is an application of the chain rule.

Step-by-Step Solution

Answer

The right-hand side of the equation in looks like the chain rule, with the important exception that it begins with $f\left(u\left(x\right)\right)$ rather than $f^{\text{'}}\left(u\left(x\right)\right)$. This is how the formula $\frac{d}{dx}{\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\text{'}}\left(x\right)$in Theorem $4.35$ is an application of the chain rule.

Step 1 . Given information

The formula $\frac{d}{dx}{\int }_a^{u\left(x\right)}f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\text{'}}\left(x\right)$ in theorem $4.35$.

Step 2 . The right-hand side of the equation in Theorem 4.35 looks like the chain rule.

With the exception that it begins with $f\left(u\left(x\right)\right)$ rather than $f^{\text{'}}\left(u\left(x\right)\right)$.

In fact, it is the chain rule, and $f\left(x\right)$ is the derivative of the area accumulation function $F\left(x\right)={\int }_a^{x}f\left(t\right)dt$.

The proof involves recognizing ${\int }_a^{u\left(x\right)}f\left(t\right)dt$as a composition and then applying the chain rule. If $F\left(x\right)={\int }_a^{x}f\left(t\right)dt$, then ${\int }_a^{u\left(x\right)}f\left(t\right)dt$ is a composition of $F\left(u\right(x\left)\right)$.By the Second Fundamental Theorem we know that $F^{\text{'}}\left(x\right)=f\left(x\right)$. Thus, by the chain rule we have,

$$\begin{gathered} \frac{d}{dx}{\int }_a^{u\left(x\right)}f\left(t\right)dt=\frac{d}{dx}\left(F\right(u\left(x\right)\left)\right) \\ =F^{\text{'}}\left(u\right(x\left)\right)u^{\text{'}}\left(x\right) \\ =f\left(u\right(x\left)\right)u^{\text{'}}\left(x\right) \end{gathered}$$