$\displaystyle \frac{d}{dx}(sec^{-1}x) = \frac{1}{|x| \sqrt{x^2-1}}$

The $\displaystyle 2$ in the original expression is a constant and can be multiplied to the identity written above. 

When dealing with the derivative of $\displaystyle sec^{-1}x$, it is important to keep the standalone $\displaystyle x$ in the denominator in absolute value bars and to make sure there is a $\displaystyle -1$ under the radical.

$\displaystyle f(x)=cot^{-1}(4x)$

First, take the derivative of the function

$\displaystyle f'(x)= -\frac{4}{1+(4x)^2}= -\frac{4}{1+16x^2}$

Especially when given inverse trigonometry derivative questions, be on the lookout for multiple functions embedded in the same problem. For example, in this problem there is both an outer function ($\displaystyle cot^{-1}$) and an inner function ($\displaystyle 4x$). Because there is more than one function, chain rule must be applied; thus, the derivative of the inner function must be multiplied to the derivative of the outer function. 

To reach the final answer, $\displaystyle f'(x)$must be evaluated at $\displaystyle x=2$.

$\displaystyle f'(2)= -\frac{4}{1+16(2)^2}= -\frac{4}{65}$

$\displaystyle \frac{d}{dx}(sin^{-1}(3x^2))=(\frac{1}{(\sqrt{1-3x^2)^2}})(6x)$

This problem requires chain rule; there is an outer function ($\displaystyle sin^{-1}$) and an inner function ($\displaystyle 3x^2$).

To simplify it further, square the $\displaystyle 3x^2$ and multiply the $\displaystyle 6x$.

$\displaystyle \frac{d}{dx}(sin^{-1}(3x^2))=(\frac{1}{(\sqrt{1-3x^2)^2}})(6x) = \frac{6x}{\sqrt{1-9x^4}}$

$\displaystyle \frac{d}{dx}(6cos^{-1}(2x))=(6)(2)(-\frac{1}{\sqrt{1-(2x)^2}})$

This problem requires chain rule, because there is an outer and inner function.

The outer function is $\displaystyle cos^{-1}$ and the inner is $\displaystyle 2x$. 

To fully simplify the expression, multiply the constants and square the $\displaystyle 2x$ term. 

$\displaystyle \frac{d}{dx}(6cos^{-1}(2x))=(6)(2)(-\frac{1}{\sqrt{1-(2x)^2}}) = \frac{-12}{\sqrt{1-4x^2}}$

$\displaystyle f(x)=(2x)tan^{-1}(x)$

This is an example of when product rule must be used to find the derivative, because there are two functions ($\displaystyle 2x$ and $\displaystyle tan^{-1}(x))$ being multiplied together within $\displaystyle f(x)$.

$\displaystyle f'(x) =(2)tan^{-1}(x)+2x(\frac{1}{1+x^2}) =2tan^{-1}(x)+\frac{2x}{1+x^2}$

$\displaystyle \frac{d}{dx}(csc^{-1}(10x))$

By examining the inverse trigonometric function derivatives, it can be shown that $\displaystyle \frac{d}{dx}(csc^{-1}x)=\frac{d}{dx}(-sec^{-1}x)$. The derivatives of these two inverse functions are negatives of one another.

$\displaystyle f(x)=(sin^{-1}x)^3$

$\displaystyle f'(x)=3(sin^{-1}x)^2*(\frac{1}{\sqrt{1-x^2}})=\frac{3(sin^{-1}x)^2}{\sqrt{1-x^2}}$

This question invokes the use of chain rule.

$\displaystyle f(x)=cos^{-1}(\sqrt{3x-1})$

This question requires the use of chain rule. In this case, there are three functions to consider: first, $\displaystyle cos^{-1}$; second, the square root; and third, $\displaystyle 3x-1$. In order to evaluate, the derivative of each of the three functions must be found and multiplied together.

$\displaystyle f'(x)=\frac{-1}{\sqrt{1-(\sqrt{3x-1})^2}}*(\frac{1}{2})*(3x-1)^{-\frac{1}{2}}*3=\frac{-3}{2(\sqrt{3x-1})(\sqrt{1-3x+1})}=\frac{-3}{2(\sqrt{3x-1})(\sqrt{2-3x})}$

This problem requires use of the chain rule twice; the first time addresses the inner function of $\displaystyle e^{2x}$ while the second accounts for the function $\displaystyle 2x$. 

$\displaystyle \frac{d}{dx}(tan^{-1}(e^{2x}))=\frac{1}{1+(e^{2x})^2}*e^{2x}*2=\frac{2e^{2x}}{1+e^{4x}}$