#### To determine

**To find:**

The derivative of the function using part 1) of the Fundamental Theorem of Calculus

#### Answer

9x+61+(3x+2)3

#### Explanation

**1) Concept:**

2) The Fundamental Theorem of Calculus: Suppose f is continuous on [a, b] then

If hs=∫auftdt, then h'=fu

3) Chain rule : Let Fx=fgx, if g is differentiable at x and f is differentiable at g(x) then F'x=f'gx·g'(x)

4) In Leibnitz notation if y=f(u) and u=g(x) are both differentiable functions then

dydx=dydu·dudx

**2) Given:**

**y=∫13x+2t1+t3dt**

3) **Calculation:**

Here

y=∫13x+2t1+t3dt Let u=3x+2 and substitute it in the above equation

Take derivative of both sides with respect to x. Thus we have

y'=ddx∫1ut1+t3dt

Using chain rule

y'=ddu∫1ut1+t3dt·dudx

Use Fundamental Theorem of Calculus part 1) with ft=t1+t3 to get

y'=u1+u3·dudx

But u=3x+2

Differentiate u

dudx=3

Substitute u=3x+2 and dudx=3 in y'

y'=3x+21+(3x+2)3·3

Simplifying

y'=9x+61+(3x+2)3

Therefore,y'=9x+61+(3x+2)3

**Conclusion:**

Therefore,y'=9x+61+(3x+2)3