If $y={{\sin }^{-1}}\left( 3x \right)+{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$, find $\dfrac{dy}{dx}$?

Hint: Differentiate both the sides of the given equation with respect to x and use the chain rule of derivative to differentiate it. Use the formulas $\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and $\dfrac{d\left( {{\sec }^{-1}}x \right)}{dx}=\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}$ to simplify. Substitute 3x in place of x in ${{\sin }^{-1}}x$ and $\dfrac{1}{3x}$ in place of x in ${{\sec }^{-1}}x$. Use the $\dfrac{u}{v}$ rule of derivative given as $\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ to find the derivative of $\dfrac{1}{3x}$. Simplify the derivative expression to get the answer.

Complete step by step answer:
Here we have been provided with the equation $y={{\sin }^{-1}}\left( 3x \right)+{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$ and we are asked to find the value of $\dfrac{dy}{dx}$. That means we have to differentiate the given equation, so differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\sin }^{-1}}\left( 3x \right) \right]}{dx}+\dfrac{d\left[ {{\sec }^{-1}}\left( \dfrac{1}{3x} \right) \right]}{dx}$
Using the chain rule of derivative we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\sin }^{-1}}\left( 3x \right) \right]}{d\left( 3x \right)}\times \dfrac{d\left( 3x \right)}{dx}+\dfrac{d\left[ {{\sec }^{-1}}\left( \dfrac{1}{3x} \right) \right]}{d\left( \dfrac{1}{3x} \right)}\times \dfrac{d\left( \dfrac{1}{3x} \right)}{dx}$
Using the formulas $\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and $\dfrac{d\left( {{\sec }^{-1}}x \right)}{dx}=\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}$ we get,
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{\left( 3x \right)}^{2}}}}\times 3+\dfrac{1}{\left| \dfrac{1}{3x} \right|\sqrt{{{\left( \dfrac{1}{3x} \right)}^{2}}-1}}\times \dfrac{d\left( \dfrac{1}{3x} \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}+\dfrac{3\left| x \right|}{\sqrt{\dfrac{1}{9{{x}^{2}}}-1}}\times \dfrac{d\left( \dfrac{1}{3x} \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}+\dfrac{3\left| x \right|\times 3\left| x \right|}{\sqrt{1-9{{x}^{2}}}}\times \dfrac{d\left( \dfrac{1}{3x} \right)}{dx} \\
\end{align}\]
We can write $\left| x \right|\times \left| x \right|=\left| {{x}^{2}} \right|$ and since $\left( {{x}^{2}} \right)$ will always be positive so we can remove the modulus function, so we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}+\dfrac{9{{x}^{2}}}{\sqrt{1-9{{x}^{2}}}}\times \dfrac{d\left( \dfrac{1}{3x} \right)}{dx}\]
Now, to differentiate the expression $\dfrac{1}{3x}$ we have to use the $\dfrac{u}{v}$ rule of derivative given as $\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. Therefore assuming u = 1 and v = 3x we get,
$\Rightarrow \dfrac{d\left( \dfrac{1}{3x} \right)}{dx}=\dfrac{3x\dfrac{d\left( 1 \right)}{dx}-1\dfrac{d\left( 3x \right)}{dx}}{{{\left( 3x \right)}^{2}}}$
We know that the derivative of a constant is 0, so we get,
$\Rightarrow \dfrac{d\left( \dfrac{1}{3x} \right)}{dx}=\dfrac{-3}{9{{x}^{2}}}$
Substituting the above value in the relation of $\dfrac{dy}{dx}$ we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}+\dfrac{9{{x}^{2}}}{\sqrt{1-9{{x}^{2}}}}\times \dfrac{-3}{9{{x}^{2}}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}-\dfrac{3}{\sqrt{1-9{{x}^{2}}}} \\
 & \therefore \dfrac{dy}{dx}=0 \\
\end{align}$
Hence, the derivative of the given expression is equal to 0.

Note: There is an alternative method also to solve the question where we have to consider two cases. In the first case we can consider x > 0 and therefore we can use the conversion ${{\sec }^{-1}}\left( \dfrac{1}{3x} \right)={{\cos }^{-1}}\left( 3x \right)$ and then apply the formula ${{\cos }^{-1}}3x+{{\sin }^{-1}}3x=\dfrac{\pi }{2}$ to get the derivative of a constant equal to 0. In the second case we have to consider x < 0 where we cannot directly use the conversion ${{\sec }^{-1}}\left( \dfrac{1}{3x} \right)={{\cos }^{-1}}\left( 3x \right)$ but first we have to make the argument positive. Therefore, the expression can be written as ${{\sin }^{-1}}\left( 3x \right)+{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)={{\sin }^{-1}}\left[ -\left( -3x \right) \right]+{{\sec }^{-1}}\left[ -\left( \dfrac{-1}{3x} \right) \right]$. Using the formulas ${{\sin }^{-1}}\left( -a \right)=-{{\sin }^{-1}}a$ and ${{\sec }^{-1}}\left( -a \right)=\pi -{{\sec }^{-1}}\left( a \right)$ we can write the simplified form of the expression as $-{{\sin }^{-1}}\left( -3x \right)+\pi -{{\sec }^{-1}}\left( \dfrac{-1}{3x} \right)$. Now since –3x > 0, we can use the similar formulas that we used in the first case to get the answer. In both the cases we will get the derivative equal to 0.