To find the derivative of a function in this form, we must make use of the chain rule, which states that for a give function: 

\(\displaystyle F(x)=f(g(x))\)

Its derivative is defined as:

\(\displaystyle F'(x)=f'(g(x))*g'(x)\)

In this case we have the function:

\(\displaystyle F(x)=sin^2(x^3)=[sin(x^3)]^2\)

which is in the form of the composite function

\(\displaystyle F(x)=f(g(h(x)))\)

in which

\(\displaystyle f(x)=x^2\) ,     \(\displaystyle g(x)=sinx\),     \(\displaystyle h(x)=x^3\)

By similar logic, the derivative of the composite function is 

\(\displaystyle F'(x)=f'(g(h(x)))*g'(h(x))*h'(x)\)

in which

\(\displaystyle f'(g(h(x)))=2sin(x^3)\)

\(\displaystyle g'(h(x))=cos(x^3)\)

and  \(\displaystyle h'(x)=3x^2\)

As a result, we have that 

\(\displaystyle F'(x)=2sin(x^3)*cos(x^3)*3x^2=6x^2sin(x^3)cos(x^3)\)

The Product Rule of derivatives states that for a given function:

\(\displaystyle f(x)=uv\)

The derivative is defined as 

\(\displaystyle f'(x)=u'v+v'u\)

In this case, for the given function

\(\displaystyle f(x)=3x^2(2x^3+10x-11)\)

\(\displaystyle u=3x^2\) , \(\displaystyle v=2x^3+10x-11\)

and the respective derivatives are:

\(\displaystyle u'=6x\) , \(\displaystyle v'=6x^2+10\)

Applying the product rule we get that

\(\displaystyle f'(x)=(6x)*(2x^3+10x-11)+(6x^2+10)*(3x^2)\)

\(\displaystyle =(12x^4+60x^2-66x)+(18x^4+30x^2)\)

\(\displaystyle =30x^4+90x^2-66x=6(5x^4+15x^2-11x)\)

To find the first derivative of this function, we must make use of the Quotient Rule of derivatives. That is, for a function

\(\displaystyle f(x)=\frac{u}{v}\)

The derivative is defined as 

\(\displaystyle f'(x)=\frac{u'v-uv'}{v^2}\) 

In this case 

\(\displaystyle u=x^2\) , \(\displaystyle v=2x+3\)

\(\displaystyle u'=2x\) , \(\displaystyle v'=2\)

As a result, 

\(\displaystyle f'(x)=\frac{(2x)(2x+3)-(2)(x^2)}{(2x+3)^2}=\frac{4x^2+6x-2x^2}{(2x+3)^2}\)

\(\displaystyle =\frac{2x^2+6x}{(2x+3)^2}=\frac{2(x^2+3x)}{(2x+3)^2}\)

First we note that 

\(\displaystyle f(-\frac{\pi}{2})=\cos(2(-\frac{\pi}{2}))+\sin(-\frac{\pi}{2})=\cos(\pi)-1=-1-1=-2\).  

Now we take the derivative  of the function to find the slope of the tangent line.

Using the chain rule we get 

\(\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)\)

\(\displaystyle f'(\theta)=-2\sin{2\theta}+\cos{\theta}\).  

Now we want to find the specific slope of the tangent line containing the \(\displaystyle \theta=-\frac{\pi}{2}\).  

We then find that the slope is 

\(\displaystyle f'(-\frac{\pi}{2})=-2\sin{(-2\frac{\pi}{2})}+\cos(\frac{-\pi}{2})=0+0=0\).  

We now plug in these values to slope-intercept form to find the y-intercept of the tangent line. 

\(\displaystyle -2=0\theta+b\) 

which gives us 

\(\displaystyle b=-2\).  

Thus the equation is simply 

\(\displaystyle y=-2\).