Taking one time derivative we get:

$\displaystyle g'(t)=\frac{d}{dt}\int_{o}^{t}e^{-s(t-b)}f(b)dt=e^{-s(t-t)}f(t)-e^{-s(t-0)}f(0)=f(t)-f(0)e^{-st}$

The derivative of the function is equal to

$\displaystyle -2\sin(2x)\tan(2x)+2\cos(2x)\sec^2(2x)$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The derivative of the function is equal to

$\displaystyle f'(x)=2x\cos(x)-x^2\sin(x)$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

To find the absolute extrema of a function on a closed interval, one must first take the first derivative of the function.

The derviatve of this function by the power rule is as follows:

$\displaystyle f'(x)=6x^2-6$

The relative extrema is when the first derivative is equal to 0, that is, there is a change in slope.

Solving for x when it is equal to zero derives:

$\displaystyle 0=6(x^2-1)=(x+1)(x-1)$

Diving by 6 and factoring gives $\displaystyle (x+1)(x-1)$or $\displaystyle x==+1,-1;$ however, since we are concerned with the interval (-2,0) our x value is -1.

We now however must find the value of f(x) at -1

$\displaystyle f(-1)=3$

$\displaystyle \begin{align*}&\text{The function }f(x)=2sin(19cos(6x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }19cos(6x)\text{ within the }2sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=19cos(6x)\\&du=-114sin(6x)\\&f'(x)=(-114sin(6x))(2cos(19cos(6x)))=-228sin(6x)cos(19cos(6x))\end{align*}$

$\displaystyle \begin{align*}&\text{Our function, }-10tan(20tan(6x))\text{, is one function nested within another.}\\&\text{Note the }20tan(6x)\text{ inside of the }-10tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=20tan(6x)\\&du=120tan(6x)^{2} + 120\\&f'(x)=(120tan(6x)^{2} + 120)(- 10tan(20tan(6x))^{2} - 10)=-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)\end{align*}$

$\displaystyle \begin{align*}&\text{The function }y=-17e^{(-19\cdot 5^{(4x)})}\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-19\cdot 5^{(4x)}\text{ is inside of the }-17e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&u=-19\cdot 5^{(4x)}\\&du=-76\cdot 5^{(4x)}ln(5)\\&\frac{dy}{dx}=(-76\cdot 5^{(4x)}ln(5))(-17e^{(-19\cdot 5^{(4x)})})=1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)\end{align*}$

$\displaystyle \begin{align*}&\text{Our function, }4^{(15cos(4x))}\text{, is one function nested within another.}\\&\text{Note the }15cos(4x)\text{ inside of the }1\cdot4^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&u=15cos(4x)\\&du=-60sin(4x)\\&f'(x)=(-60sin(4x))(4^{(15cos(4x))}ln(4))=-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)\end{align*}$

$\displaystyle \begin{align*}&\text{The function }f(x)=\frac{1}{5^{(17x^{2})}}\text{, can be seen as a function nested within another.}\\&\text{Notice the }-17x^{2}\text{ within the }1\cdot5^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&u=-17x^{2}\\&du=-34x\\&f'(x)=(-34x)(\frac{ln(5)}{5^{(17x^{2})}})=-\frac{(34xln(5))}{5^{(17x^{2})}}\end{align*}$

$\displaystyle \begin{align*}&\text{The function }y=-13ln(16cos(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }16cos(3x)\text{ is inside of the }-13ln()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&u=16cos(3x)\\&du=-48sin(3x)\\&\frac{dy}{dx}=(-48sin(3x))(-\frac{13}{(16cos(3x))})=\frac{(39sin(3x))}{cos(3x)}\end{align*}$