Question

If we are given an inverse trigonometric expression as $y={{\sin }^{-1}}\left( 3x \right)+{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$, then find the value of $\dfrac{dy}{dx}$.

Answer 1

We will use the differentiation of inverse trigonometric formulas for inverse sine and secant terms. These are given by $\dfrac{d}{dx}{{\sin }^{-1}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ where $1-{{x}^{2}}\ne 0$ and $\dfrac{d}{dx}{{\sec }^{-1}}\left( x \right)=\dfrac{1}{{{x}^{2}}\sqrt{1-\dfrac{1}{{{x}^{2}}}}}$ where $x\ne -1,0,1$. With the help of these formulas we will be able to solve the question.

Complete step-by-step answer:
Now, we will consider the inverse trigonometric expression $y={{\sin }^{-1}}\left( 3x \right)+{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$...(i). We will differentiate equation (i) with respect of x. Therefore we have $\dfrac{d}{dx}y=\dfrac{d}{dx}{{\sin }^{-1}}\left( 3x \right)+\dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$...(ii).
Now we will consider the term $\dfrac{d}{dx}{{\sin }^{-1}}\left( 3x \right)$ and find its value by using the formula $\dfrac{d}{dx}{{\sin }^{-1}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ where $1-{{x}^{2}}\ne 0$. After this we get
$\begin{aligned} & \dfrac{d}{dx}{{\sin }^{-1}}\left( 3x \right)=\dfrac{1}{\sqrt{1-{{\left( 3x \right)}^{2}}}}\times 3 \\\ & \Rightarrow \dfrac{d}{dx}{{\sin }^{-1}}\left( 3x \right)=\dfrac{3}{\sqrt{1-9{{x}^{2}}}} \\\ \end{aligned}$
Here, ${{c}_{1}}$ is any arbitrary constant. Now we will consider the term $\dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)$. At this step we will apply the formula and $\dfrac{d}{dx}{{\sec }^{-1}}\left( x \right)=\dfrac{1}{{{x}^{2}}\sqrt{1-\dfrac{1}{{{x}^{2}}}}}$ where $x\ne -1,0,1$ to evaluate it. Thus, we will get
$\dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=\dfrac{1}{{{\left( \dfrac{1}{3x} \right)}^{2}}\sqrt{1-\dfrac{1}{{{\left( \dfrac{1}{3x} \right)}^{2}}}}}\times \dfrac{d}{dx}\left( \dfrac{1}{3x} \right)$. The value of $\dfrac{d}{dx}\left( \dfrac{1}{3x} \right)$ can be carried out by solving it like so, $\dfrac{d}{dx}\left( \dfrac{1}{3x} \right)=\dfrac{1}{3}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)$. As we know that $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$. Therefore we get
$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{1}{3x} \right)=\dfrac{1}{3}\times -\dfrac{1}{{{x}^{2}}} \\\ & \Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{3x} \right)=-\dfrac{1}{3{{x}^{2}}} \\\ \end{aligned}$
After substituting this value in above equation we will have
$\begin{aligned} & \dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=\dfrac{1}{\dfrac{1}{9{{x}^{2}}}\sqrt{1-\dfrac{1}{\dfrac{1}{9{{x}^{2}}}}}}\times \left( -\dfrac{1}{3{{x}^{2}}} \right) \\\ & \Rightarrow \dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=-\dfrac{1}{\dfrac{3{{x}^{2}}}{9{{x}^{2}}}\sqrt{1-\dfrac{1}{\dfrac{1}{9{{x}^{2}}}}}} \\\ & \Rightarrow \dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=-\dfrac{1}{\dfrac{1}{3}\sqrt{1-\dfrac{1}{\dfrac{1}{9{{x}^{2}}}}}} \\\ \end{aligned}$
Here, ${{c}_{4}}$ is any arbitrary constant. Now we will use the property $\dfrac{a}{\dfrac{b}{c}}=\dfrac{ac}{b}$ in $\dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=-\dfrac{1}{\dfrac{1}{3}\sqrt{1-\dfrac{1}{\dfrac{1}{9{{x}^{2}}}}}}$ so, to get $\dfrac{d}{dx}{{\sec }^{-1}}\left( \dfrac{1}{3x} \right)=-\dfrac{3}{\sqrt{1-9{{x}^{2}}}}$. After this we are going to substitute these values in equation (ii). This results into
$\begin{aligned} & \dfrac{d}{dx}y=\dfrac{3}{\sqrt{1-9{{x}^{2}}}}+-\dfrac{3}{\sqrt{1-9{{x}^{2}}}} \\\ & \Rightarrow \dfrac{d}{dx}y=0 \\\ \end{aligned}$
Hence, the value of $\dfrac{dy}{dx}$ is 0.

Note: We can also use the basic triple angle formula of sine which is given by $\sin \left( 3x \right)=3\sin \left( x \right)-4{{\sin }^{3}}\left( x \right)$. With the help of this formula we can find out $\sin \left( 3x \right)+4{{\sin }^{3}}\left( x \right)=3\sin \left( x \right)$ and put it in the equation (i). Then we can solve it further and get a different equation. Here all the constants as c are the arbitrary constants. This means that with the presence of these in the solution will never hinder the answer. As in this question we do not have definite values this is why we are using the constants while using the differentiation operation.

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