Question

use part one of the fundamental theorem of calculus to find the derivative of the function.

$y = \int_{4}^{3x + 5} \frac{t}{1 + t^{3}} dt$

$y = $

Explanation:

Step1: Recall the fundamental theorem of calculus and chain rule

If \(y = \int_{a}^{u(x)}f(t)dt\), then \(y^\prime=f(u(x))\cdot u^\prime(x)\) by the fundamental theorem of calculus part 1 and the chain rule. Here \(a = 4\), \(u(x)=3x + 5\), and \(f(t)=\frac{t}{1 + t^{3}}\).

Step2: Find \(u^\prime(x)\)

Differentiate \(u(x)=3x + 5\) with respect to \(x\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(u^\prime(x)=(3x+5)^\prime=3\).

Step3: Find \(f(u(x))\)

Substitute \(t = u(x)=3x + 5\) into \(f(t)\). So \(f(u(x))=\frac{3x + 5}{1+(3x + 5)^{3}}\).

Step4: Calculate \(y^\prime\)

By the formula \(y^\prime=f(u(x))\cdot u^\prime(x)\), we substitute \(f(u(x))\) and \(u^\prime(x)\) we found above. \(y^\prime=\frac{3x + 5}{1+(3x + 5)^{3}}\cdot3=\frac{3(3x + 5)}{1+(3x + 5)^{3}}\).

Answer:

\(\frac{3(3x + 5)}{1+(3x + 5)^{3}}\)