Calculating Derivative of h(x) using Fundamental Theorem

[tandoorichicken]

Hello everyone, its been a while.

It's been almost 4 months since I did anything calculus related so I forgot all of my skills.

The problem is:
Use the Fundamental Theorem of Calculus to find the derivative of the function
$$h(x) = \int_{2}^{\frac{1}{x}} \arctan{t} \,dt$$

 

[vsage]

IIRC (whichI might not) the fundamental theorem of calculus says that given F(x) = S(f(x),x,a,b) F'(x) = f(b)-f(a)

 

[BLaH!]

Consider the function $$F(x) = \int_{a}^{x} f(t) \,dt$$.

The Fundamental Theorem of Calculus is given by: $$\frac{dF}{dx} = f(x)$$. In your case the upper integration limit is $$1/x$$. Therefore, you will have to use the chain rule. Let $$u=1/x \Rightarrow \frac{dh}{dx} = \frac{dh}{du}\frac{du}{dx} = -\frac{1}{x^2}arctan(\frac{1}{x})$$