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