Question

1. find the following derivatives.

a. $\frac{d}{dx}\int_{3}^{x}e^{t}\cos(t^{3})dt$
b. $\frac{d}{dx}\int_{x}^{3}e^{t}\cos(t^{3})dt$
c. $\frac{d}{dx}\int_{7x}^{x^{2}}t\sin(t)dt$

Explanation:

Step1: Use the Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus Part 1 states that if \(F(x)=\int_{a}^{x}f(t)dt\), then \(F^\prime(x) = f(x)\).

For part A

Let \(F(t)\) be an antiderivative of \(e^{t}\cos(t^{3})\), i.e., \(F^\prime(t)=e^{t}\cos(t^{3})\). Then \(\int_{3}^{x}e^{t}\cos(t^{3})dt=F(x)-F(3)\).
Differentiating with respect to \(x\) gives:
\(\frac{d}{dx}\int_{3}^{x}e^{t}\cos(t^{3})dt=\frac{d}{dx}(F(x)-F(3))\)
Since \(\frac{d}{dx}F(3) = 0\) (a constant), by the Fundamental Theorem of Calculus Part 1, we have \(e^{x}\cos(x^{3})\)

For part B

Let \(G(t)\) be an antiderivative of \(e^{t}\cos(t^{3})\), i.e., \(G^\prime(t)=e^{t}\cos(t^{3})\). Then \(\int_{x}^{3}e^{t}\cos(t^{3})dt=G(3)-G(x)\)
Differentiating with respect to \(x\) gives:
\(\frac{d}{dx}\int_{x}^{3}e^{t}\cos(t^{3})dt=\frac{d}{dx}(G(3)-G(x))\)
Since \(\frac{d}{dx}G(3) = 0\) (a constant), and using the chain - rule \(\frac{d}{dx}(-G(x))=-G^\prime(x)\), we get \(-e^{x}\cos(x^{3})\)

For part C

Let \(H(t)\) be an antiderivative of \(t\sin(t)\), i.e., \(H^\prime(t)=t\sin(t)\). Then \(\int_{7x}^{x^{2}}t\sin(t)dt=H(x^{2})-H(7x)\)
Differentiating with respect to \(x\) using the chain - rule:
\(\frac{d}{dx}(H(x^{2})-H(7x))=H^\prime(x^{2})\cdot2x - H^\prime(7x)\cdot7\)
Since \(H^\prime(t) = t\sin(t)\), we substitute \(t = x^{2}\) and \(t=7x\)
\(=x^{2}\sin(x^{2})\cdot2x-7x\sin(7x)\cdot7\)
\(=2x^{3}\sin(x^{2})-49x\sin(7x)\)

Answer:

A. \(e^{x}\cos(x^{3})\)
B. \(-e^{x}\cos(x^{3})\)
C. \(2x^{3}\sin(x^{2})-49x\sin(7x)\)