Let $y(x) = \int_0^x \frac{\sin(t)}{t}dt $ find maximums and minimums of $y(x)$.
First let $F(x) = \int_0^x \frac{\sin(t)}{t}dt$ and $f(t) = \frac{\sin(t)}{t}$ then $F'(c) = f(c) $ then if $ F'(c) = 0$ i have that $ \frac{\sin(c)}{c} = 0$ then $ t = 2 \pi n$ and $ (\frac{\sin(t)}{t})'= \frac{t \cos(t) - \sin(t)}{t^2}$ but how get the vaues of t ?.
Some help for this please i stuck here i don't know how to find minimums and maximums of $y(x)$.
Hint: You are looking for points where $\frac{\sin t}{t}=0$, no points where $(\frac{\sin t}{t})'=0$.
Let $F$-antiderivative of $f(x)=\frac{\sin x}{x}$ We are looking for points $x$ where $y'(x)=0$, so let's compute $y'$:
$y'(x)=(\int_{0}^{x} f(t) dt)'=(F(x)-F(0))'=F'(x)=f(x)=\frac{\sin x}{x}$