Question

use the fundamental theorem of calculus to find the derivative of (y = int_{x}^{ 9}\frac{cos t}{t + 7}dt)

Explanation:

Step1: Recall FTC part 1

If $y = \int_{a}^{u(x)}f(t)dt$, then $y'=f(u(x))\cdot u'(x)$ by the Fundamental Theorem of Calculus part 1. Here $a = - 6$, $u(x)=\sqrt{x}$, and $f(t)=\frac{\cos t}{t + 7}$.

Step2: Find $u'(x)$

The derivative of $u(x)=\sqrt{x}=x^{\frac{1}{2}}$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $u'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

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

Substitute $t = u(x)=\sqrt{x}$ into $f(t)$. So $f(u(x))=\frac{\cos\sqrt{x}}{\sqrt{x}+7}$.

Step4: Calculate $y'$

By the chain - rule from the Fundamental Theorem of Calculus, $y'=f(u(x))\cdot u'(x)$. Then $y'=\frac{\cos\sqrt{x}}{\sqrt{x}+7}\cdot\frac{1}{2\sqrt{x}}=\frac{\cos\sqrt{x}}{2\sqrt{x}(\sqrt{x}+7)}$.

Answer:

$\frac{\cos\sqrt{x}}{2\sqrt{x}(\sqrt{x}+7)}$