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 du/dx using u = 1/x, we need to apply the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).
In this case, we have h(u) = integral from 1 to u of arctan(2t)dt and u = 1/x. We want to find dh/dx, which is the derivative of h with respect to x.
Using the chain rule, we have:
dh/dx = dh/du * du/dx
Let's break this down further:
1. dh/du is the derivative of h with respect to u. From Part 1, we found h'(u) = arctan(2). Therefore, dh/du = arctan(2).
2. du/dx is the derivative of u with respect to x. Since u = 1/x, we can rewrite u as u = x^(-1). When we differentiate this, we apply the power rule, which states that if we have a function f(x) = x^n, its derivative is given by f'(x) = n*x^(n-1).
Applying the power rule, we have:
du/dx = d/dx (x^(-1))
= -x^(-2)
= -1/x^2
Now, we can substitute these values into the formula dh/dx = dh/du * du/dx:
dh/dx = arctan(2) * (-1/x^2)
Therefore, the derivative of h with respect to x, dh/dx, is equal to arctan(2) multiplied by -1 divided by x^2.