what is the derivative with respect to x of the integral with respect to t of sin(t)/t with a lower limit of 1 and an upper limit of x^2
meaning
d/dx(1 INT x^2 (sin(t)/t)dt)
F(x) = ∫[1,x^2] f(t) dt
dF/dx = f(x^2) * 2x = sin(x^2)/x^2 * 2x = 2/x sin(x^2)
This is just the chain rule in disguise.
If dF/dt = f(t) then
∫[a,b] f(t) dt = F(b)-F(a)
Now, if b is a function of x (in this case, x^2), then
d/dx F(x) = dF/db db/dx - dF/da * da/dx
= f(x^2) * 2x - 0 since da/dx = 0
Nani?
To find the derivative with respect to x of the given integral, we can apply the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if we have a function F(x) defined as the integral of another function f(t) with respect to t, from a constant a to x, then the derivative of F(x) with respect to x is equal to f(x).
In this case, we have:
F(x) = ∫[1, x^2] (sin(t)/t) dt
To find the derivative of F(x), we can differentiate the integral with respect to x, keeping in mind the limits of integration.
So, dF(x)/dx = d/dx ∫[1, x^2] (sin(t)/t) dt
Now, to apply the Fundamental Theorem of Calculus, we need to find the antiderivative of (sin(t)/t) with respect to t.
Let's call this antiderivative function G(t). So, G'(t) = (sin(t)/t).
Next, we substitute the limits of integration into G(t):
G(x^2) - G(1)
Now, we can use the Chain Rule to find the derivative of G(x^2) and G(1) with respect to x.
dG(x^2)/dx = G'(x^2) * d(x^2)/dx = (sin(x^2)/(x^2)) * 2x = 2x * sin(x^2)/x^2 = 2sin(x^2)/x
dG(1)/dx = G'(1) * d(1)/dx = sin(1)/1 = sin(1)
Finally, we substitute these results back into our expression:
dF(x)/dx = d/dx ∫[1, x^2] (sin(t)/t) dt = dG(x^2)/dx - dG(1)/dx
= 2sin(x^2)/x - sin(1)
Therefore, the derivative with respect to x of the given integral is 2sin(x^2)/x - sin(1).