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
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)
2 answers
Nani?