#### To determine

**To find:**

A formula for ddx∫g(x)h(x)ftdt

#### Answer

ddx∫g(x)h(x)ftdt=fhx·h'x-f(gx)·g'(x)

#### Explanation

**1) Concept:**

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

∫abfxdx=Fb-F(a), where F is antiderivative of f, that is F'=f.

**2) Calculations:**

If Ft is the antiderivative of f(t) then by Fundamental Theorem of Calculus ∫g(x)h(x)ftdt=Fhx-F(gx)

By taking derivative on both sides, the above equation becomes

ddx∫gxhxftdt=ddx[Fhx-F(gx)]

By applying the chain rule, theabove equation becomes

ddx∫g(x)h(x)ftdt=F'hx·h'x-F'(gx)·g'(x)

Substituting F'=f in the above equation, it becomes

ddx∫g(x)h(x)ftdt=fhx·h'x-f(gx)·g'(x)

**Conclusion:**

Hence, the formula for ddx∫g(x)h(x)ftdt is

ddx∫g(x)h(x)ftdt=fhx·h'x-f(gx)·g'(x)