Let A(x)=∫x^(2),∞; (e^-t/t)dt for x>0.
Find A'(x)
To find A'(x), we can use the Fundamental Theorem of Calculus. Recall that the Fundamental Theorem of Calculus states that if F(x) is a continuous function on the interval [a, b] and f(x) is its derivative, then
∫[a, b] f(x) dx = F(b) - F(a).
In our case, A(x) is defined as ∫x^(2),∞ (e^(-t)/t) dt for x > 0.
To find A'(x), we need to differentiate A(x) with respect to x. However, A(x) is an integral with a variable upper limit, which complicates the differentiation process. We can employ a technique called the Leibniz Integral Rule, also known as Differentiation under the Integral Sign, to overcome this challenge.
The Leibniz Integral Rule states that if a function F(x, t) is continuous on an open rectangle R: a < x < b, α(t) < t < β(t), and the partial derivatives ∂F/∂x and ∂F/∂t exist and are continuous over R, then
d/dx ∫[α(x), β(x)] F(x, t) dt = ∫[α(x), β(x)] (∂F/∂x)(x, t) dt + (∂F/∂t)(x, β(x)) β'(x) - (∂F/∂t)(x, α(x)) α'(x),
where α(x) and β(x) are functions of x.
In our case, F(x, t) = e^(-t)/t, α(x) = x^2, and β(x) = ∞. Notice that we have α(x), β(x), (∂F/∂x)(x, t), and (∂F/∂t)(x, t) all given explicitly. Therefore, we can calculate A'(x) using the Leibniz Integral Rule.
Let's differentiate A(x) using the Leibniz Integral Rule:
d/dx ∫[α(x), β(x)] F(x, t) dt = ∫[α(x), β(x)] (∂F/∂x)(x, t) dt + (∂F/∂t)(x, β(x)) β'(x) - (∂F/∂t)(x, α(x)) α'(x).
Applying the Leibniz Integral Rule, we have:
A'(x) = (∂/∂x)∫[x^2, ∞] (e^(-t)/t) dt + (∂/∂t)(e^(-t)/t)(x, ∞) * (∞)' - (∂/∂t)(e^(-t)/t)(x, x^2) * (x^2)'
Now, we need to compute the partial derivatives (∂F/∂x)(x, t) and (∂F/∂t)(x, t):
(∂F/∂x)(x, t) = 0, since there is no x dependency in F(x, t).
(∂F/∂t)(x, t) = -e^(-t) / t^2 - e^(-t) / t.
We can now substitute these values back into the equation:
A'(x) = 0 + (-e^(-∞) / ∞^2 - e^(-∞) / ∞) * (∞)' - (-e^(-x^2) / (x^2)^2 - e^(-x^2) / x^2) * (x^2)'
As x approaches infinity, e^(-∞) and (∞)' are both zero, so we can simplify our equation:
A'(x) = 0 + 0 - (-e^(-x^2) / (x^2)^2 - e^(-x^2) / x^2) * (x^2)'
Simplifying further:
A'(x) = e^(-x^2) / x^2.
Therefore, the derivative of A(x) with respect to x (A'(x)) is e^(-x^2) / x^2.