#### To determine

**To find:**

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

#### Answer

4x3·cos2(x4)

#### Explanation

**1) Concept:**

i. The Fundamental Theorem of Calculus-Suppose f is continuous on [a, b] then,

if hs=∫auftdt, then h'=fu

ii. 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)

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

dydx=dydu·dudx

**2) Given:**

**y=∫0x4cos2θdθ**

3) **Calculation:**

Here,

**y=∫0x4cos2θdθ**

Let u=x4 and substitute it in the above equation

Take derivative of both sides

y'=ddx∫0ucos2θdθ

Using chain rule

y'=ddu∫0ucos2θdθ·dudx

Use Fundamental Theorem of Calculus part 1) with fθ=cos2θ to obtain

ddu∫0ucos2θdθ=cos2u Thus

y'=cos2u·dudx

But u=x4

Differentiate u

dudx=4x3

Substitute u=x4 and dudx=4x3 in y' to get

y'=cos2(x4)·4x3

Therefore, y'=4x3·cos2(x4)

**Conclusion:**

Therefore,y'=4x3·cos2(x4)