Let F(x)= ∫ √(t^2+21) . Find F'(10) and F''(10) . **the bounds on the integral are 10 up to x. **

Answer 1

The derivative and integral are essentially opposites and in this case kind of "cancel" each other out.

The derivative of an integral of a function is just the original function (this time of x), provided there is just an x there in the upper bound.

In this case, F'(x) = f(x), so F'(10) = f(10) = √10² +21 = √121 = 11.

Also, F''(x) = d/dx F'(x) = d/dx f(x) = f'(x).

In this case, we're going to have to use the chain rule to evaluate the derivative.

f'(x) = d/dx (√ (x²+21)) = d/dx (x²+21)^1/2 = (1/(2*(x²+21)^1/2))*2x

= 2x/2(x²+21)^1/2 = x/√(x²+21)

F''(10) = f'(10) = 10/√(10²+21) = 10/√(100+21) = 10/√121 = 10/11.