Question

Differentiate the value \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\] with respect to \[{{\cos }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] when $x\ne 0$

Answer 1

Hint: To solve this question, first of all assume variable for given inverse trigonometric function then use the fact that the derivative of p with respect to q, where, p and q are both function of t is given by $\dfrac{dp}{dq}=\dfrac{\dfrac{dp}{dt}}{\dfrac{dq}{dt}}$ Also, we will use the several trigonometric formulas and chain rule of differentiation given as below:
\[\begin{align}
  & \Rightarrow \text{si}{{\text{n}}^{2}}\theta =1-\text{co}{{\text{s}}^{2}}\theta \\
 & \Rightarrow \dfrac{\text{sin}\theta }{\text{cos}\theta }=\text{tan}\theta \\
 & \Rightarrow 2\text{sin}\theta \text{cos}\theta \text{=sin2}\theta \\
 & \Rightarrow \text{sin}\theta =\text{cos}\left( \dfrac{\pi }{2}-\theta \right) \\
\end{align}\]
Chain rule of differentiation when f(x) and g(x) are function of x \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)\]

Complete step-by-step solution:
Let us assume some variables for the given terms.
Let \[u={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\text{ and v=}{{\cos }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]
We have to differentiate u with respect to v. Then, we will apply basic differentiation rule of du and dv which is given as
\[\dfrac{du}{dv}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
So, firstly to calculate $\dfrac{du}{dv}$ we will separately calculate $\dfrac{du}{dx}\text{ and }\dfrac{dv}{dx}$
That is we will differentiate u with respect to x and v with respect to x.
We have \[u={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\]
Differentiating both sides with respect to x we get:
\[\dfrac{du}{dx}=\dfrac{d}{dx}={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\]
To do so, let us assume x = cost in above
\[\dfrac{du}{dx}=\dfrac{d}{dx}\left( \text{ta}{{\text{n}}^{-1}}\left( \dfrac{\sqrt{1-{{\cos }^{2}}t}}{\text{cos t}} \right) \right)\]
We know a trigonometric identity as \[\begin{align}
  & \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1 \\
 & \Rightarrow \text{si}{{\text{n}}^{2}}\theta =1-\text{co}{{\text{s}}^{2}}\theta \\
\end{align}\]
Taking square root both sides \[\text{sin}\theta \text{=}\sqrt{1-\text{co}{{\text{s}}^{2}}\theta }\]
Using this above by taking $\theta =t$ we get
\[\dfrac{du}{dx}=\dfrac{d}{dx}\left( \text{ta}{{\text{n}}^{-1}}\left( \dfrac{\text{sin t}}{\text{cos t}} \right) \right)\]
Now, we know that \[\dfrac{\text{sin}\theta }{\text{cos}\theta }=\text{tan}\theta \]
Using this in above by taking $\theta =t$ we get
\[\begin{align}
  & \dfrac{du}{dx}=\dfrac{d}{dx}\left( \text{ta}{{\text{n}}^{-1}}\left( \text{tan t} \right) \right) \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \text{t} \right)\text{ as }\left( \text{ta}{{\text{n}}^{-1}}\left( \text{tan }\theta \right) \right)=\theta \\
\end{align}\]
So, finally we have u as a function of t. Then, applying chain rule of differentiation which states that, chain rule of differentiation where f(x) and g(x) are function of x.
\[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)\]
In above we get
\[\Rightarrow \dfrac{du}{dx}=\dfrac{du}{dt}\times \dfrac{dt}{dx}\]
We have x = cos t
Differentiate both side with respect to t and using $\dfrac{d}{d\theta }\text{cos}\theta =-\text{sin}\theta $ in above we get:
\[\begin{align}
  & \Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( \text{cos t} \right)=-\text{sin t} \\
 & \Rightarrow \dfrac{dx}{dt}=-\text{sin t} \\
\end{align}\]
Reversing the above
\[\Rightarrow \dfrac{dt}{dx}=\dfrac{-1}{\text{sin t}}\]
Then, finally we have \[\Rightarrow \dfrac{du}{dx}=\dfrac{du}{dt}\times \dfrac{dt}{dx}\]
Substituting all values obtained above
\[\begin{align}
  & \Rightarrow \dfrac{du}{dx}=1\left( \dfrac{-1}{\text{sin t}} \right) \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{-1}{\text{sin t}} \\
\end{align}\]
Using relation stated before that \[\text{sin t} \text{=}\sqrt{1-\text{co}{{\text{s}}^{2}}t}\] in above
\[\begin{align}
  & \Rightarrow \dfrac{du}{dx}=\dfrac{-1}{\sqrt{1-\text{co}{{\text{s}}^{2}}t }} \\
 & \text{as x=cos t} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
So, we have obtained $\dfrac{du}{dx}$
Similarly we will obtain $\dfrac{dv}{dx}$
\[v=\text{co}{{\text{s}}^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\]
Let x = cos t
Then, substituting this value of x in v we get:
\[\Rightarrow v=\text{co}{{\text{s}}^{-1}}\left( 2\text{cos t}\sqrt{1-\text{co}{{\text{s}}^{2}}t} \right)\]
Using identity stated above as \[\text{sin}\theta \text{=}\sqrt{1-\text{co}{{\text{s}}^{2}}\theta }\] we get
\[\Rightarrow v=\text{co}{{\text{s}}^{-1}}\left( 2\text{cos t sin t} \right)\]
Now, we have a trigonometric identity given as
\[\Rightarrow \text{2sin}\theta \text{cos}\theta =\text{sin}2\theta \]
Using this above $\theta =t$ we get
\[\Rightarrow v=\text{co}{{\text{s}}^{-1}}\left( \text{sin2t} \right)\]
We have a relation between $\sin \theta \text{ and cos}\theta $ as
\[\Rightarrow \text{sin}\theta =\text{cos}\left( \dfrac{\pi }{2}-\theta \right)\]
Using this above and $\theta =2t$ we get
\[\begin{align}
  & \Rightarrow v={{\cos }^{-1}}\left( \text{cos}\left( \dfrac{\pi }{2}-2t \right) \right) \\
 & \Rightarrow v=\dfrac{\pi }{2}-2t \\
\end{align}\]
So, we have obtained v as a function of t, then, $\dfrac{dv}{dx}$ can be obtained by applying chain rule of differentiating stated above
\[\Rightarrow \dfrac{dv}{dx}=\dfrac{dv}{dt}\times \dfrac{dt}{dx}\]
We have \[\Rightarrow v=\dfrac{\pi }{2}-2t\]
Differentiating above with respect to we get:
\[\dfrac{dv}{dt}=0-2=-2\]
And we already had \[\Rightarrow \dfrac{dt}{dx}=\dfrac{-1}{\text{sin t}}\]
Using \[\begin{align}
  & \text{sin}\theta \text{=}\sqrt{1-\text{co}{{\text{s}}^{2}}\theta } \\
 & \Rightarrow \dfrac{dt}{dx}=\dfrac{-1}{\sqrt{1-\text{co}{{\text{s}}^{2}}\theta }} \\
\end{align}\]
and value of cos t = x
\[\Rightarrow \dfrac{dt}{dx}=\dfrac{-1}{\sqrt{1-{{\text{x}}^{2}}}}\]
Now, substituting value of $\dfrac{dt}{dx}\text{ and }\dfrac{dv}{dt}$ we get
\[\begin{align}
  & \Rightarrow \dfrac{dv}{dx}=\dfrac{dv}{dt}\times \dfrac{dt}{dx} \\
 & \Rightarrow \dfrac{dv}{dx}=\dfrac{-2\left( -1 \right)}{\sqrt{1-{{x}^{2}}}} \\
 & \Rightarrow \dfrac{dv}{dx}=\dfrac{\left( -2 \right)\left( -1 \right)}{\sqrt{1-{{x}^{2}}}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\
\end{align}\]
Substituting value of equation (ii) and (iii) in equation (i) we get
\[\begin{align}
  & \dfrac{du}{dv}=\dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} \\
 & \dfrac{du}{dv}=\dfrac{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}{\dfrac{-2\left( -1 \right)}{\sqrt{1-{{x}^{2}}}}} \\
\end{align}\]
Cancelling the common term $\dfrac{-1}{\sqrt{1-{{x}^{2}}}}$ we get:
\[\dfrac{du}{dv}=\dfrac{-1}{2}=-0.5\]
So, the derivative of \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\text{ and }{{\cos }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] when \[x\ne 0\text{ is -0}\text{.5}\Rightarrow \dfrac{-1}{2}\]

Note: While proceeding at the steps of solution of such type of question where complex functions involving inverse trigonometric functions is given like here \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)\text{ and }{{\cos }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)\] is given, go for cancelling the ${{\tan }^{-1}}$ term by trying to obtain $\tan \theta $ inside of ${{\tan }^{-1}}$ so that we can have ${{\tan }^{-1}}\left( \tan \theta \right)=\theta $ then, differentiation becomes easy. Similarly, in ${{\cos }^{-1}}$ term try to obtain cos inside ${{\cos }^{-1}}$ to get ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $ to make differentiation easy.