Solving for the derivative using the above rules and the chain rule.

\[\frac{d}{dx}\left[\cos^{-1}\left(\sqrt{x^{3}+2}\right)\right]=\frac{-1}{\sqrt{1-\left(\sqrt{x^{3}+2}\right)^{2}}}\cdot\frac{d}{dx}\left[\left(x^{3}+2\right)^{-\frac{1}{2}}\right]=\] \[\frac{-1}{\sqrt{1-x^{3}-2}}\cdot\frac{1}{2\sqrt{x^{3}+2}}\cdot3x^{2}=\] \[\frac{-3x^{2}}{2\sqrt{x^{3}+2}\sqrt{-x^{3}-1}}\]

Solving for the derivative using, the product rule, the above rules, the exponential and logarithmic rules, and the chain rule.

\[\frac{d}{dx}\left[3\ln\left(x\right)\tan^{-1}\left(x^{2}+\sqrt{x}\right)\right]=\] \[3\left[\left(\frac{d}{dx}\left[\ln\left(x\right)\right]\right)\tan^{-1}\left(x^{2}+\sqrt{x}\right)+\ln\left(x\right)\cdot\frac{d}{dx}\left[\tan^{-1}\left(x^{2}+\sqrt{x}\right)\right]\right]=\] \[3\left[\left(\frac{1}{x}\cdot\tan^{-1}\left(x^{2}+\sqrt{x}\right)\right)+\frac{\ln\left(x\right)}{1+\left(x^{2}+\sqrt{x}\right)^{2}}\cdot\frac{d}{dx}\left[x^{2}+\sqrt{x}\right]\right]=\] \[3\left[\frac{\tan^{-1}\left(x^{2}+\sqrt{x}\right)}{x}+\frac{\ln\left(x\right)}{1+x^{4}+2x^{2}\sqrt{x}+x}\cdot\left(2x+\frac{1}{2\sqrt{x}}\right)\right]=\] \[\frac{3\tan^{-1}\left(x^{2}+\sqrt{x}\right)}{x}+\frac{3\ln\left(x\right)\left(2x+\frac{1}{2\sqrt{x}}\right)}{x^{4}+2x^{2}\sqrt{x}+x+1}\]

To do this, we must first rearrange the function so that it's in terms of \(x\).

1. Rearrange Function \[x=4\sin^{2}\left(y\right)\] \[\frac{x}{4}=\sin^{2}\left(y\right)\] \[\sqrt{\frac{x}{4}}=\sin\left(y\right)\] \[y=\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)\]
2. Find the Derivative \[y'=\frac{1}{\sqrt{1-\left(\frac{\sqrt{x}}{2}\right)^{2}}}\cdot\frac{d}{dx}\left[\frac{\sqrt{x}}{2}\right]=\] \[\frac{1}{\sqrt{1-\frac{x}{4}}}\cdot\frac{1}{4\sqrt{x}}=\frac{1}{4\sqrt{x}\sqrt{1-\frac{x}{4}}}\]

1. Find the derivative \[\frac{d}{dx}\left[\csc^{-1}\left(x^{3}\right)\right]=\] \[\frac{-1}{\left|x^{3}\right|\sqrt{\left(x^{3}\right)^{2}-1}}\cdot\frac{d}{dx}\left[x^{3}\right]=\] \[\frac{-3x^{2}}{\left|x^{3}\right|\sqrt{x^{6}-1}}\]
2. Find the where the is equal to \(0\) and/or is discontinuous \[\] To determine where rational functions, like the one above, do not exist, we need to find where the denominator is equal to \(0\) and/or is discontinuous. \[\] Let's first examine \(\left|x^{3}\right|\). \(\left|x^{3}\right|\) is only equal to \(0\) at \(x=0\); therefore, at \(x=0\) the derivative function does not exist. \[\] Next, let's examine \(\sqrt{x^{6}-1}\). If \(x\) were equal to \(1\) or \(-1\), the square root function would be equal to \(0\); therefore at; therefore, at \(x=1 \mbox{ and } x=-1\) the derivative function does not exist. Also, the square root function can not be equal to a negative value; therefore, \(x
3. Find the general domain on which the derivative does not exist \[\] The derivative does not exist at the points \(x=-1\), \(x=0\), and \(x=1\). Also, the derivative function must be in less than \(-1\) and greater than \(1\). \[\therefore\] The derivative function exists when \(x<-1 \mbox{="" and="" }="" x="">1\). \[\{x|x<-1 \cap="" x="">1\}\] \[\mbox{ or }\] \[(-\infty,-1) \cap (1,\infty)\]</-1></-1>

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If possible, check with a graphing calculator.

1. Find the derivative \[\frac{d}{dx}\left[\cot^{-1}\left(\ln\left(x\right)\right)\right]=\]\[\frac{-1}{1+\ln^{2}\left(x\right)}\cdot\frac{d}{dx}\left[\ln\left(x\right)\right]=\] \[\frac{-1}{1+\ln^{2}\left(x\right)}\cdot\frac{1}{x}=-\frac{1}{x+x\ln^{2}\left(x\right)}\]
2. Find the where the is equal to \(0\) and/or is discontinuous \[\] Let's first factor the denominator. \(\frac{-1}{x\left(1+\ln^{2}\left(x\right)\right)}\). If we look at the first of the two functions, \(x\) is equal to \(0\) at \(x=0\); therefore, at \(x=0\) the derivative function does not exist. The second function, \(1+\ln^{2}\left(x\right)\), only exists when \(x>0\) since the natural log function can not be equal to \(0\)
3. Find the general domain on which the derivative does not exist \[\] The derivative does not exist at \(x=0\), and it must also be greater the \(0\). \[\therefore\] \[\{x|x>0\}\] \[\mbox{ or }\] \[(0, \infty)\]

\[\]

If possible, check with a graphing calculator.

A simple way to solve this is to substitute \(x=\frac{3}{5}\) into the derivative of each of the options.

1. \(\mbox{A.}\) \[f'\left(x\right)=\frac{-1}{\sqrt{1-x^{2}}}\] \[\] \[f'\left(\frac{3}{5}\right)=\frac{-1}{\sqrt{1-\left(\frac{3}{5}\right)^{2}}}=\frac{-1}{\sqrt{\frac{25}{25}-\frac{9}{25}}}=\] \[-\frac{1}{\sqrt{\frac{16}{25}}}=-\frac{1}{\frac{4}{5}}=-\frac{5}{4}\] \[\] As we can see, \(\mbox{A.}\) is not the correct answer. \[\]
2. \(\mbox{B.}\) \[f'\left(x\right)=\frac{1}{1+x^{2}}\] \[\] \[f'\left(\frac{3}{5}\right)=\frac{1}{1+\left(\frac{3}{5}\right)^{2}}=\frac{1}{\frac{25}{25}+\frac{9}{25}}=\frac{1}{\frac{34}{25}}=\frac{25}{34}\] \[\] As we can see, \(\mbox{B.}\) is not the correct answer. \[\]
3. \(\mbox{C.}\) \[\] Since the derivative of \(\cot^{-1}\left(x\right)\) will yield the same result as \(\tan^{-1}\left(x\right)\), except negative, \(f'(\frac{3}{5})\) for \(\cot^{-1}\left(x\right)\) will be the negative result of \(f'(\frac{3}{5})\) for \(\tan^{-1}\left(x\right)\); therefore, \(\mbox{C.}\) is not the correct answer.\[\]
4. \(\mbox{D.}\) \[\] Since \(\sin^{-1}\left(x\right)\) will yield the same result as \(\cos^{-1}\left(x\right)\), except positive, \(\mbox{D.}\) is the correct answer. \[\] \[f'\left(\frac{3}{5}\right)=\frac{5}{4}\] \[\]
5. \(\mbox{E.}\) \[f'\left(x\right)=\frac{-1}{\left|x\right|\sqrt{x^{2}-1}}\] \[\] \[f'\left(\frac{3}{5}\right)=\frac{-1}{\left|\frac{3}{5}\right|\sqrt{\frac{9}{25}-1}}=\frac{-1}{\left|\frac{3}{5}\right|\sqrt{-\frac{16}{25}}}=\frac{-1}{\left|\frac{3}{5}\right|\sqrt{\frac{16}{25}}i}\] As we can see, \(\mbox{E.}\) is not the correct answer. \[\]

\(\mbox{D.}\) is the correct answer.

1. Find the value of the function at \(x=4\) \[f\left(4\right)=\sec^{-1}\left(\sqrt{4}\right)=1.0471\]
2. Find the derivative \[f'\left(x\right)=\frac{1}{\sqrt{x}\sqrt{x-1}}\cdot\frac{d}{dx}\left[\sqrt{x}\right]=\] \[\frac{1}{\sqrt{x}\sqrt{x-1}}\cdot\frac{1}{2\sqrt{x}}=\] \[\frac{1}{2\sqrt{x}\sqrt{x}\sqrt{x-1}}=\frac{1}{2x\sqrt{x-1}}\]
3. Evaluate the derivative at \(x=4\) \[f'\left(4\right)=\frac{1}{8\sqrt{4-1}}=\frac{1}{8\sqrt{3}}\]
4. Write the equation of the tangent line \[y-\sec^{-1}\left(2\right)=\frac{1}{8\sqrt{3}}\left(x-4\right)\] \[\mbox{ or }\] \[y=\frac{1}{8\sqrt{3}}x-\frac{1}{2\sqrt{3}}+\sec^{-1}\left(2\right)\]