Let A(x)=∫x^(2),∞, ((e^(-t)/(t))dt for x>0. Find A'(x).
To find the derivative of A(x), we'll use the Fundamental Theorem of Calculus. According to this theorem, if we have a function F(x) defined as the integral of another function f(t), p(t), q(t), where p(t) and q(t) are the limits, then the derivative of F(x) with respect to x is given by:
F'(x) = f(q(x)) * q'(x) - f(p(x)) * p'(x)
For our given function A(x), we have:
A(x) = ∫x^(2),∞, ((e^(-t)/(t))dt
Let's assign f(t) = e^(-t)/t, p(t) = x^2, and q(t) = ∞. Since the upper limit is infinity, we'll need an extra step to deal with it. Let's do that:
A(x) = ∫x^(2),∞, ((e^(-t)/(t))dt
= ∫x^(2),K, ((e^(-t)/(t))dt, where K is any finite positive number.
Now, we'll find the derivative of A(x) by applying the Fundamental Theorem of Calculus:
A'(x) = d/dx [∫x^(2),K, ((e^(-t)/(t))dt]
= (∂/∂x) [∫x^(2),K, ((e^(-t)/(t))dt]
= f(q(x)) * q'(x) - f(p(x)) * p'(x)
Applying the derivative to the integral limits, we obtain
A'(x) = f(K) * 0 - f(x^2) * 2x
Recall that f(t) = e^(-t)/t, so we have:
A'(x) = -(e^(-x^2)/x) * 2x
= -2e^(-x^2)
Therefore, the derivative of A(x) is A'(x) = -2e^(-x^2).