Question

The derivative of ${{\sec }^{-1}}\left( \dfrac{1}{2{{x}^{2}}-1} \right)$ with respect to $\sqrt{1-{{x}^{2}}}$ at $x=\dfrac{1}{2}$ is
A. $-4$
B. 4
C. 2
D. $-2$

Answer 1

Hint: We first define the chain rule and how the differentiation of composite function works. We differentiate the main function with respect to the intermediate function and then differentiation of the intermediate function with respect to $x$. We take multiplication of these two different differentiated values.

Complete step-by-step solution:
We need to find the derivative of ${{\sec }^{-1}}\left( \dfrac{1}{2{{x}^{2}}-1} \right)$ with respect to $\sqrt{1-{{x}^{2}}}$ at $x=\dfrac{1}{2}$ using chain rule.
We assume $x=\cos \alpha $ and by putting the value we get $2{{x}^{2}}-1=2{{\cos }^{2}}\alpha -1=\cos 2\alpha $.
So, ${{\sec }^{-1}}\left( \dfrac{1}{\cos 2\alpha } \right)={{\sec }^{-1}}\left( \sec 2\alpha \right)=2\alpha $.
From inverse law we get $\alpha ={{\cos }^{-1}}x$. So, $2\alpha =2{{\cos }^{-1}}x$.
Here we assume the function is $m\left( x \right)=2{{\cos }^{-1}}x$ and the other function is $n\left( x \right)=\sqrt{1-{{x}^{2}}}$.
We need to find ${{\left[ \dfrac{dm}{dn} \right]}_{x=\dfrac{1}{2}}}$. We express it as $\dfrac{dm}{dn}=\dfrac{dm}{dx}\times \dfrac{1}{\dfrac{dn}{dx}}$.
\[\dfrac{dm}{dx}=\dfrac{d}{dx}\left[ 2{{\cos }^{-1}}x \right]=-\dfrac{2}{\sqrt{1-{{x}^{2}}}}\]
\[\dfrac{dn}{dx}=\dfrac{d}{dx}\left[ \sqrt{1-{{x}^{2}}} \right]=\dfrac{-2x}{2\sqrt{1-{{x}^{2}}}}=-\dfrac{x}{\sqrt{1-{{x}^{2}}}}\]
We place the values of the differentiations and get
\[\begin{align}
  & \dfrac{dm}{dn} \\
 & =\dfrac{dm}{dx}\times \dfrac{1}{\dfrac{dn}{dx}} \\
 & =\left( -\dfrac{2}{\sqrt{1-{{x}^{2}}}} \right)\times \left( \dfrac{1}{-\dfrac{x}{\sqrt{1-{{x}^{2}}}}} \right) \\
 & =\dfrac{2\sqrt{1-{{x}^{2}}}}{x\sqrt{1-{{x}^{2}}}} \\
 & =\dfrac{2}{x} \\
\end{align}\]
Now the value of ${{\left[ \dfrac{dm}{dn} \right]}_{x=\dfrac{1}{2}}}$ will be \[{{\left[ \dfrac{2}{x} \right]}_{x=\dfrac{1}{2}}}=\dfrac{2}{\dfrac{1}{2}}=4\]. The correct option is B.

Note: We need remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t cancelling out the part \[d\left[ h\left( x \right) \right]\]. Canceling the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.