Even though we are working with trigonometric functions, the rules of derivatives remain the same.  Since the graph of a derivative function is the graph of the rate of change of the original function, then when the graph of the derivative function is negative, the original function is decreasing.  Even when the derivative is increasing, if it is negative the original function is still decreasing, just by less of a factor.  When the derivative function is at zero, then the original function is constant.  When the derivative function is positive, then the original function is increasing.

We need to use the Chain Rule to take both the derivative of the trigonometric function and the quantity within the trig function.

$\displaystyle f(x) = cot(5x^2)$

$\displaystyle f'(x) = csc^2(5x^2) * \frac{d}{dx}[5x^2]$

$\displaystyle f'(x) = 10xcsc^2(5x^2)$

Recall that the derivative of the tangent function is $\displaystyle sec(x)$.  So we need to take the derivative of the exponent of $\displaystyle tan^2(x)$ and we also need the derivative of the actual tangent function.

$\displaystyle f(x) = tan^2(x)$

$\displaystyle f'(x) = 2tan(x) *\frac{d}{dx}[tan(x)]$

$\displaystyle f'(x) = 2sec^2(x)tan(x)$

The inverse of a function can be found by substituting yvariables for the $\displaystyle x$ variables found in the function, then setting the function equal to $\displaystyle x$. By next isolating $\displaystyle y$, the inverse function is written. Then, the notation $\displaystyle f^{-1}$is used to describe the newly written function as being the inverse of the original function. The answer choice “$\displaystyle f^{-1}(x) =\frac{1}{2}x+2$” is correct.

This question asks you to recognize the correct notation of a differentiating inverse functions problem. First, it is key to recognize that the equation needs to have the same variable throughout, thus eliminating the answer choices $\displaystyle (f^{-1})'(y) = \frac{1}{f'(f^{-1}(x)) }$and $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(y))}$. Next, there should be no constants in the correct equation; thus, $\displaystyle (f^{-1})'(x) = \frac{2}{f'(f^{-1}(x))}$ is incorrect. The correct choice is $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

Let $\displaystyle f (x) = x^5+2x+1$

It is important to recognize the relationship between a function and its inverse to solve.

If $\displaystyle f (0) = (0)^5+2(0)+1=1$, solving for the inverse function will produce $\displaystyle f^{-1}(1)=0$.

To find the derivative of an inverse function, use: $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$

$\displaystyle (f^{-1})'(1) = \frac{1}{f'(f^{-1}(1))}=\frac{1}{f' (0)}=\frac{1}{5(0)^4+ 2}=\frac{1}{0 + 2}=\frac{1}{2}$

Therefore, $\displaystyle (f^{-1})'(1)=\frac{1}{2}$

To find the derivative of the inverse of $\displaystyle f (x)$, it is useful to first solve for $\displaystyle f^{-1}$.

This will help because $\displaystyle f^{-1}$is needed in the derivative equation, $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

$\displaystyle f (x) =x^6 \rightarrow f^{-1} (x) =\sqrt[6]{x}$

Next, the equation for the derivative of an inverse function can be evaluated.

$\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[6]{x})}$

After substituting in $\displaystyle \sqrt[6]{x}$ for $\displaystyle f^{-1} (x)$, the derivative of $\displaystyle f (x),$ or $\displaystyle f'(x) =6x^5$, is applied:

$\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[6]{x})}=\frac{1}{6(\sqrt[6]{x})^5}=\frac{1}{6x^5_6}$

Therefore, the correct answer is $\displaystyle (f^{-1})'(x)=\frac{1}{6x^5_6}$

To find the derivative of the inverse of $\displaystyle f (x)$, it is useful to first solve for $\displaystyle f^{-1}$.

This will help because $\displaystyle f^{-1}$ is needed in the derivative equation, $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

$\displaystyle f (x) =e^x \rightarrow f^{-1} (x) =ln(x)$

Next, the equation for the derivative of an inverse function can be evaluated.

$\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(ln(x))}$

After substituting in $\displaystyle ln(x) for f^{-1} (x)$, the derivative of $\displaystyle f (x)$, or$\displaystyle f'(x) =e^x$, is applied:

$\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(ln(x))}=\frac{1}{e^{ln(x)}}=\frac{1}{x}$

Therefore, the correct answer is $\displaystyle (f^{-1})'(x)=\frac{1}{x}$

To find the derivative of the inverse of $\displaystyle f (x)$, it is useful to first solve for $\displaystyle f^{-1}$.

This will help because $\displaystyle f^{-1}$is needed in the derivative equation, $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

$\displaystyle f (x)=\sqrt[3]{x} \rightarrow f^{-1} (x)=x^3$

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in $\displaystyle x^3$ for $\displaystyle f^{-1} (x)$, the derivative of $\displaystyle f (x)$, or $\displaystyle f'(x)=(\frac{1}{3})(x^{-\frac{2}{3}})$, is applied:

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

Therefore, the correct answer is $\displaystyle (f^{-1})'(x)=3x^2$.

To find the derivative of the inverse of $\displaystyle f (x)$, it is useful to first solve for $\displaystyle f^{-1}$.

This will help because $\displaystyle f^{-1}$is needed in the derivative equation, $\displaystyle (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

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

Next, the equation for the derivative of an inverse function can be evaluated.

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

After substituting in $\displaystyle 1-\frac{1}{2}x$ for $\displaystyle f^{-1} (x)$, the derivative of $\displaystyle f (x)$, or $\displaystyle f'(x)= -2$, is applied:

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

Therefore, the correct answer is $\displaystyle (f^{-1})'(x)=-\frac{1}{2}$.