Im trying to solve the following question:
The Derivative of the integral of (sin(t))/t dt with the upper limit of arcsin(x) and the lower limit of 1.
As far as I understand its easy to find the derivative of integral if the upper limit is larger than the lower limit and equal to x. But now I have a function as the upper limit. How can you solve that?
In its most general form, the Fundamental Theorem of Calculus (Analysis) states that for an appropriate integrand,
$$ \frac{d}{dx}\int_{u(x)} ^{v(x)} f(t) \ dt = f(v(x))v'(x) - f(u(x))u'(x) $$
So for your integral,
$$ \frac{d}{dx} \int_1 ^{\arcsin x} \frac{\sin t}{t} \ dt = \frac{\sin{(\arcsin x)}}{\arcsin x} \frac{d}{dx} (\arcsin x) $$
Can you finish?