h(x)= integral from (1, 1/x) arctan(2t)dt

part 1: let U= 1/x and du= ?

-> using u=1/x, we can write h(x)= integral from (1, 1/x) arctan (2t)dt as h(u)= integral from (1,u) arctan(2t)dt and h'(u)= arctan (2)

Part 2: By the chain Rule, for functions h(u) and u(x), we have:

-> dh/dx= ______ du/dx

To find dh/dx using the chain rule, we need to find du/dx.

From Part 1, we have u = 1/x. To find du/dx, we can differentiate both sides of this equation with respect to x:

du/dx = d(1/x)/dx

To differentiate 1/x with respect to x, we can rewrite it as x^(-1):

du/dx = d(x^(-1))/dx

Applying the power rule for differentiation, where d/dx(x^n) = nx^(n-1):

du/dx = (-1)x^(-1-1)

du/dx = -x^(-2)

So, du/dx = -1/x^2.

Now, we can substitute this value into the chain rule formula:

dh/dx = (dh/du) * (du/dx)

From Part 1, we found dh/du = arctan(2).

Substituting du/dx = -1/x^2, we get:

dh/dx = arctan(2) * (-1/x^2)

Therefore, dh/dx = -arctan(2)/x^2.