Derivative of Integral -

Let $$h(x)=\int_2^{10/x}arctan(t)dt$$. Use the fundamental Theorem of Calculus to find $$h'(x)$$.

So I believe the answer is $$arctan(10/x)-arctan(2)$$, but am being told it is incorrect. Any other ideas?

My thought process was that the derivative of an integral will give back the integrand, then evaluate $$F(b)-F(a)$$ where F is the original integrand.

Thanks!

• Apply the Fundamental Theorem of Calculus (and apply chain rule) – Morgan Rodgers Jan 24 at 0:43

if $$F(x) - F(b) = \int_b^{x} f(t) \ dt$$

$$F'(x) = f(x)$$

This is the fundamental theorem of calculus.

We have

$$F(\frac {10}{x}) - F(b) = \int_b^{\frac{10}{x}} f(t) \ dt$$

$$\frac {d}{dx}\left(F(\frac {10}{x}) - F(b)\right) = \frac {d}{dx}\int_b^{\frac{10}{x}} f(t) \ dt$$

Remember, b is a constant, so $$\frac {d}{dx} F(b) = 0$$

For the other term we are going to use the chain rule.

$$F'(\frac {10}{x}) = f(\frac {10}{x})(\frac {d}{dx} \frac {10}{x}) = f(\frac{10}{x})(\frac {-10}{x^2})$$

$$-\frac {10\arctan \frac {10}{x}}{x^2}$$

It's true that if $$F(x)=\int_a^x f(t)\,dt$$, then $$F'(x)=f(x)$$. But in your case you have $$h(x)=F(\tfrac{10}x),$$ where $$F(x)=\int_2^x\arctan(t)\,dt$$. So to find $$h'$$ you need to apply the Chain Rule and the Fundamental Theorem of Calculus.