Use part I of the Fundamental Theorem of Calculus to find the derivative of:

Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of:

h(x) = ∫ (top of the integral symbol says "sin(x)" and bottom of integral symbol says "-3") then followed by the equation (cos(t^5)+t)dt

h'(x) = ??

Answer 1

So:

Let's talk about what the derivative would be if, instead of "sin(x)", the upper limit of integration were "x". If that were the problem, then we would just exchange the "t" with "x" and h'(x) would be cos(x⁵ + x) because the derivative of an integral is just the function inside (part I of the Fundamental Theorem of Calculus).

But, this has a more complicated upper limit of integration, so, by the chain rule, we must multiply by the derivative of the upper limit of integration. We must also "plug in" the upper limit of integration. So, that makes it:

h'(x) = cos(sin⁵(x) + sin(x))•(sin(x))'

h'(x) = cos(sin⁵(x) + sin(x))•cos(x)

Notice we get to ignore the lower limit of integration. The reason for that is that it is a constant and the derivative of a constant is zero.