How do you use fundamental method of calculus to find the derivative of the function $y=\int{{{\sin }^{3}}t}dt$ from $\left[ {{e}^{x}},0 \right]$.

Hint: We know that the derivative of $\int\limits_{g\left( x \right)}^{h\left( x \right)}{f\left( t \right)dt}$ with respect to x is equal to f( h (x)) h’ (x) –f(g(x)) g’(x) where h’(x) is the derivative h(x) with respect to x and g’ (x) is the derivative of g(x) with respect to x. We can use this formula to solve the given question.

Complete step-by-step solution:
We have to find the derivative of $y=\int{{{\sin }^{3}}t}dt$ where the limits are from ${{e}^{x}}$ to 0.
We know that derivative of $\int\limits_{g\left( x \right)}^{h\left( x \right)}{f\left( t \right)dt}$ with respect to x is equal to f( h (x)) h’ (x) –f(g(x)) g’(x) , So here function f is ${{\sin }^{3}}t$ , function g is ${{e}^{x}}$ and function h is 0.
So, if we will apply the formula, we will get
$\dfrac{d}{dx}\left( \int\limits_{{{e}^{x}}}^{0}{{{\sin }^{3}}tdt} \right)={{\sin }^{3}}\left( 0 \right)\times 0-{{\sin }^{3}}\left( {{e}^{x}} \right)\dfrac{d{{e}^{x}}}{dx}$
We know the derivative of ${{e}^{x}}$ with respect to x is equal to ${{e}^{x}}$ and derivative of any constant number is equal to 0.
So we can write $\dfrac{d}{dx}\left( \int\limits_{{{e}^{x}}}^{0}{{{\sin }^{3}}tdt} \right)=-{{e}^{x}}{{\sin }^{3}}\left( {{e}^{x}} \right)$
$-{{e}^{x}}{{\sin }^{3}}\left( {{e}^{x}} \right)$ is the correct answer of $\dfrac{d}{dx}\left( \int\limits_{{{e}^{x}}}^{0}{{{\sin }^{3}}tdt} \right)$

Note: While applying the formula $\int\limits_{g\left( x \right)}^{h\left( x \right)}{f\left( t \right)dt}$ = f( h (x)) h’ (x) –f(g(x)) g’(x) keep in mind the function f does not have variable x within it , if the function f is f ( x, t) then we can not apply the above formula. We should note that the limits of the integration should be of the same variable to the variable with respect to which we will differentiate.