Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

h'(x) =

I got -arctan x dx/x^2..think it's wrong

2. Originally Posted by Jgirl689
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

h'(x) =

I got -arctan x dx/x^2..think it's wrong
If $h(x)=\int_a^{g(x)} f(t) dt$, then:

$h'(x)=f\left(g(x)\right) g'(x)$.

In your answer, It should be $arctan \left( \frac{1}{x} \right)$ not just $arctan(x)$.

3. Originally Posted by Jgirl689
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

h'(x) =

I got -arctan x dx/x^2..think it's wrong
if $h(x) = \int_a^u f(t) \, dt$ , then $h'(x) = f(u) \cdot \frac{du}{dx}$

$h'(x) = \arctan\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$

4. Let $\arctan t = F'(t)$

Then $h(x) = \int _6 ^{1/x} \arctan t dt = F(1/x) - F(6)$

$h'(x) = F'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$