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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

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Last updated date: 29th May 2024
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Answer
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Hint: The differentiation of composite functions containing inverse and some algebraic function can be solved by some suitable substitution, which we can find by observing the algebraic function. Here by observation, we can see that \[x=\,\cos \theta \] will work as it is clearly visible that on putting \[x=\,\cos \theta \], \[2{{x}^{2}}-1\] will change to \[x=\,\cos 2\theta \].

Complete step by step solution:
For a function \[f(x)\] finding \[f'(x)\] or \[\dfrac{dy}{dx}\] at a given point is called finding the derivative of \[f(x)\] at that point. \[\dfrac{dy}{dx}\] or \[f'(x)\] represents the instantaneous rate of change of \[y\] w.r.t \[x\].
Here in the given question, we have to find the derivative of a function w.r.t another function.
mathematically:
Let \[y\,=\,f(x)\] ; \[z\,=\,g(x)\] then \[\dfrac{dy}{dz}\,\,=\,\,\dfrac{dy/dx}{dz/dx}\,\,\,=\,\,\,\dfrac{f'(x)}{g'(x)}\].
Here we have \[y\,=\,\,{{\sec }^{-1}}\left( \dfrac{1}{2{{x}^{2}}-1} \right)\] and \[z\,=\,\sqrt{1-{{x}^{2}}}\].
to find: \[\dfrac{dy}{dz}\]
Let’s solve \[\dfrac{dy}{dx}\] and \[\dfrac{dz}{dx}\] separately.
First \[\dfrac{dy}{dx}\]:
It will be quite convenient if we simplify the given inverse function instead of directly differentiating it using the chain rule.
By observing we can see that we should put \[x=\,\cos \theta \]. Therefore \[y\] becomes
\[y\,=\,\,\,{{\sec }^{-1}}\left( \dfrac{1}{2{{\cos }^{2}}\theta -1} \right)\]
As we know that \[\cos 2\theta \,\,=\,\,{{\cos }^{2}}\theta \,-\,{{\sin }^{2}}\theta \,=\,\,2{{\cos }^{2}}\theta -1\]
Hence \[y\,=\,\,\,{{\sec }^{-1}}\left( \dfrac{1}{2{{\cos }^{2}}\theta -1} \right)\,\,\,=\,\,\,{{\sec }^{-1}}\left( \dfrac{1}{\cos 2\theta } \right)\,\,=\,\,{{\sec }^{-1}}\left( \sec 2\theta \right)\,\,=\,\,2\theta \,\]
As \[x=\,\cos \theta \] alternatively \[\theta \,\,=\,\,{{\cos }^{-1}}x\]
\[\Rightarrow \,\,\,y\,\,=\,\,2\theta \,\,=\,\,2{{\cos }^{-1}}x\].
\[\therefore \,\,\,\,\dfrac{dy}{dx}\,\,\,=\,\,\,\dfrac{d}{dx}(2{{\cos }^{-1}}x)\,\,=\,\,2\times \dfrac{-1}{\sqrt{1-{{x}^{2}}}}\,\,\,=\,\,\,\dfrac{-2}{\sqrt{1-{{x}^{2}}}}\] \[\left\{ \because \,\dfrac{d}{dx}({{\cos }^{-1}}x)\,\,=\,\,\dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right\}\]
Hence \[\dfrac{dy}{dx}\,=\,\dfrac{-2}{\sqrt{1-{{x}^{2}}}}\]
Now \[\dfrac{dz}{dx}\]:
This function can be differentiated by using the chain rule of differentiation.
As \[z\,=\,\sqrt{1-{{x}^{2}}}\]
\[\Rightarrow \,\,\dfrac{dz}{dx}\,=\,\dfrac{d}{dx}(\sqrt{1-{{x}^{2}}})\,=\,\dfrac{1}{\sqrt{1-{{x}^{2}}}}\times \,(0-2x)\,=\,\,\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}\]
Now as we know that \[\dfrac{dy}{dz}\,\,=\,\,\dfrac{dy/dx}{dz/dx}\,\,\,=\,\,\,\dfrac{f'(x)}{g'(x)}\]
Hence, \[\dfrac{dy}{dz}\,\,\,=\,\,\,\dfrac{\dfrac{-2}{\sqrt{1-{{x}^{2}}}}}{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\,\,\,=\,\,\,\dfrac{-2}{\sqrt{1-{{x}^{2}}}}\times \,\dfrac{\sqrt{1-{{x}^{2}}}}{-2x}\,\,\,=\,\,\,\dfrac{1}{x}\]
Now we need to find the value of \[\dfrac{dy}{dz}\] at \[x=\dfrac{1}{2}\]
\[\therefore \,\,{{\left( \dfrac{dy}{dz} \right)}_{x=\dfrac{1}{2}}}\,\,=\,\,\,\dfrac{1}{\left( \dfrac{1}{2} \right)}\,\,=\,\,2\]
Hence, 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 equals to \[2\].

Note:
As inverse trigonometric functions are defined for a fixed domain so while operating with inverse trigonometric functions we should take care of its domain.
In the function \[{{\sec }^{-1}}(\sec x)\], if the domain is violating then suitable addition or subtracting must be done so that in \[{{\sec }^{-1}}(\sec x)\]its input i.e.,\[x\in \left[ 0,\pi \right]\,-\,\left\{ \dfrac{\pi }{2} \right\}\]
So that we can write \[{{\sec }^{-1}}(\sec x)\,\,=\,\,x\].