How do I compute $\int_{-x^2}^{0}f(t)dt$ derivative ratio $x$? If 
$$ 
f(t)=\begin{cases}\frac{\sin t}{t}  &  t\neq 0\\\\1 & t=0,\end{cases}
$$ 
then, how do I compute the following derivative ratio $x$?
The desired function is:

$$\int_{-x^2}^{0}f(t)dt$$

 A: It is correct however as a general suggestion you often don't want to write the symbol $0'$ to indicate that you will derivate a constant. Here you are using the fundamental theorem of calculus in combination with the chain rule. 
First of all, the value of a function at $t=0$ is not important since Lebesgue Riemann-integrability criterion asserts that as long as the points of discontinuity has "measure zero" (if you are not familiarize with "measures" you just have to know that any finite or countable set has measure zero) the function is Riemann-integrable. Now, provided that the funcion is integrable the fundamental theorem of calculus says that
$$ \frac{d}{dx}\int_a^xf(t)  \; dt = f(x)$$
for any constant $a$. Two things are different in your problem: you have an $x^2$ instead of a $x$ and the integration limits are interchanged. The second thing can be solved by multyplying the whole expression by a minus sign and the first one by means of the chain rule, which combined with the fundamental theorem states that if $u=u(x)$ is a function of $x$ then
$$ \frac{d}{dx} \int_a^{u(x)}f(t) \; dt = \left( \frac{d}{du} \int_a^{u}f(t)\right)\left(\frac{du}{dx}\right) = f(u(x)) \cdot \frac{du}{dx} $$
Finally by multiplying by a minus sign and by taking $u(x)=-x^2$ and therefore $u'(x)=-2x$ you get your answer without having to do any calculations.
A: I managed to solve my question and may be better to write it here.
$\color{green}{Answer}$:
$$\left ( \int_{-x^2}^{0}f(t)dt \right )'=0'f(0)-(-x^2)'-f(-x^2)=2xf(-x^2)=2x(\frac{1}{-x^2})\sin(-x^2)=\frac{2\sin x^2}{x}$$
Thank you! :)
