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

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:

& \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{aligned}$$ Chain rule of differentiation when f(x) and g(x) are function of x $$\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{aligned} & \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{aligned}$$ Taking square root both sides $$\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{aligned} & \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{aligned}$$ So, finally we have u as a function of t. Then, applying chain rule of differentiation which states that, chain rule of differentiation where f(x) and g(x) are function of x. $$\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{aligned} & \Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( \text{cos t} \right)=-\text{sin t} \\\ & \Rightarrow \dfrac{dx}{dt}=-\text{sin t} \\\ \end{aligned}$$ Reversing the above $$\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{aligned} & \Rightarrow \dfrac{du}{dx}=1\left( \dfrac{-1}{\text{sin t}} \right) \\\ & \Rightarrow \dfrac{du}{dx}=\dfrac{-1}{\text{sin t}} \\\ \end{aligned}$$ Using relation stated before that $$\text{sin t} \text{=}\sqrt{1-\text{co}{{\text{s}}^{2}}t}$$ in above $$\begin{aligned} & \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{aligned}$$ So, we have obtained $\dfrac{du}{dx}$ Similarly we will obtain $\dfrac{dv}{dx}$ $$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{aligned} & \Rightarrow v={{\cos }^{-1}}\left( \text{cos}\left( \dfrac{\pi }{2}-2t \right) \right) \\\ & \Rightarrow v=\dfrac{\pi }{2}-2t \\\ \end{aligned}$$ So, we have obtained v as a function of t, then, $\dfrac{dv}{dx}$ can be obtained by applying chain rule of differentiating stated above $$\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{aligned} & \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{aligned}$$ and value of cos t = x $$\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{aligned} & \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{aligned}$$ Substituting value of equation (ii) and (iii) in equation (i) we get $$\begin{aligned} & \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{aligned}$$ Cancelling the common term $\dfrac{-1}{\sqrt{1-{{x}^{2}}}}$ we get: $$\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.