Let F(x) = ∫ from a to u(x) f(t) dt, where u is differentiable. Then F'(x) equals what?

Let F(x) = ∫ from a to u(x) f(t) dt, where u is differentiable. Then F'(x) equals what?

• A. f(x) • u'(x)
• B. f(u(x)) • u'(x)
• C. f'(u(x)) • x'
• D. ∫ f(t) dt

Answer: (B)

When you differentiate an integral whose upper limit depends on x, you apply the Fundamental Theorem of Calculus together with the chain rule. If F(x) = ∫ from a to u(x) f(t) dt, then differentiate by first regarding the integral as a function of its upper limit: G(v) = ∫ from a to v f(t) dt, so G'(v) = f(v). Since F(x) = G(u(x)), by the chain rule F'(x) = G'(u(x)) · u'(x) = f(u(x)) · u'(x). This requires f to be continuous enough for the FTC to apply. So the derivative is f(u(x)) multiplied by u'(x). The other forms don’t fit because the integrand is evaluated at the upper limit, not at x, and there’s no need for f′. The integral itself isn’t the derivative.

When you differentiate an integral whose upper limit depends on x, you apply the Fundamental Theorem of Calculus together with the chain rule. If F(x) = ∫ from a to u(x) f(t) dt, then differentiate by first regarding the integral as a function of its upper limit: G(v) = ∫ from a to v f(t) dt, so G'(v) = f(v). Since F(x) = G(u(x)), by the chain rule F'(x) = G'(u(x)) · u'(x) = f(u(x)) · u'(x). This requires f to be continuous enough for the FTC to apply.

So the derivative is f(u(x)) multiplied by u'(x). The other forms don’t fit because the integrand is evaluated at the upper limit, not at x, and there’s no need for f′. The integral itself isn’t the derivative.