#### To determine

**To find:**

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

#### Answer

x2(x2+1)

#### 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 chain rule reads; if y=f(u) and u=g(x) are both differentiable functions then

dydx=dydu·dudx

**2) Given:**

**hx=∫1xz2z4+1dz**

3) **Calculation:**

Given that

hx=∫1xz2z4+1dz

Let u=x and substitute it in the above equation

Take derivative of both sides

h'x=ddx∫1uz2z4+1dz

Using chain rule

h'x=ddu∫1uz2z4+1dz·dudx

Here,

Use fundamental theorem of Calculus part 1) with fz=z2z4+1 to find ddu∫1uz2z4+1dz Thus we have ddu∫1uz2z4+1dz=u2u4+1

h'x=u2u4+1·dudx

But u=x

Differentiating u with respect to x we have

dudx=12x

Substitute u=x and dudx=12x in h'x to get

h'x=(x)2(x)4+1·12x

Simplify

h'x=xx2+1·12x

Simplify

h'x=x2(x2+1)

Therefore,h'(x)=x2(x2+1)

**Conclusion:**

Therefore,h'(x)=x2(x2+1)