Let F(x) = displaystyleint0^x √t^2+1\,dt. Find F'(x).

Derivative of an accumulation function

Let $F(x) = \displaystyle\int_0^x \sqrt{t^2+1}\,dt$ . Find $F'(x)$ .

Hint

Apply the Fundamental Theorem of Calculus, Part 1: if $F(x) = \int_a^x f(t)\,dt$ with $f$ continuous, then $F'(x) = f(x)$ . Here the upper limit is simply $x$ , so no chain rule is needed.

Step-by-step solution

The function $f(t) = \sqrt{t^2+1}$ is continuous everywhere, so the Fundamental Theorem of Calculus Part 1 applies directly to $F(x) = \int_0^x f(t)\,dt$ .

FTC Part 1 states that $F'(x) = f(x)$ . Substituting $x$ in place of $t$ in the integrand gives $F'(x) = \sqrt{x^2+1}$ .

Derivative of an accumulation function

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