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Question: Differentiate \[\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right...

Differentiate [tan⁡−1(xa2−x2)]\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right], with respect to x.

Explanation

Solution

Hint- Here, we will proceed by simplifying the given function (which needs to be differentiated) with the help of a proper substitution.

Let the function be y=[tan⁡−1(xa2−x2)]y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]
Put x=asin⁡θx = a\sin \theta in the above function, we get
⇒y=[tan⁡−1(asin⁡θa2−(asin⁡θ)2)]=[tan⁡−1(asin⁡θa2[1−(sin⁡θ)2])]=[tan⁡−1(asin⁡θa1−(sin⁡θ)2)]\Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2}\left[ {1 - {{\left( {\sin \theta } \right)}^2}} \right]} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}} \right)} \right]
As, we know that (sin⁡θ)2+(cos⁡θ)2=1⇒1−(sin⁡θ)2=(cos⁡θ)2{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \Rightarrow 1 - {\left( {\sin \theta } \right)^2} = {\left( {\cos \theta } \right)^2}
Using this above identity to simplify the function whose differentiation needs to be carried out

⇒y=[tan⁡−1(asin⁡θa1−(sin⁡θ)2)]=[tan⁡−1(sin⁡θ(cos⁡θ)2)]=[tan⁡−1(sin⁡θcos⁡θ)]=[tan⁡−1(tan⁡θ)]=θ ⇒y=θ →(1)  \Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {{{\left( {\cos \theta } \right)}^2}} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\tan \theta } \right)} \right] = \theta \\\ \Rightarrow y = \theta {\text{ }} \to {\text{(1)}} \\\

Also, we know that x=asin⁡θ⇒sin⁡θ=xa⇒θ=sin⁡−1(xa) →(2)x = a\sin \theta \Rightarrow \sin \theta = \dfrac{x}{a} \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(2)}}
Substituting the value obtained from equation (2) in equation (1), we get
⇒y=sin⁡−1(xa) →(3)\Rightarrow y = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(3)}}
Also we know that d[sin⁡−1(f(x))]dx=[11−(f(x))2]×[d(f(x))dx]\dfrac{{d\left[ {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right]}}{{dx}} = \left[ {\dfrac{1}{{\sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} }}} \right] \times \left[ {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}} \right]
Differentiating both sides of equation (3) with respect to xx, we get

⇒dydx=d[sin⁡−1(xa)]dx = [11−(xa)2]×[d(xa)dx]=[1(a2−x2a2)]×[1a]=[aa2−x2]×[1a] ⇒dydx=1a2−x2  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left[ {{{\sin }^{ - 1}}\left( {\dfrac{x}{a}} \right)} \right]}}{{dx}}{\text{ = }}\left[ {\dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{x}{a}} \right)}^2}} }}} \right] \times \left[ {\dfrac{{d\left( {\dfrac{x}{a}} \right)}}{{dx}}} \right] = \left[ {\dfrac{1}{{\sqrt {\left( {\dfrac{{{a^2} - {x^2}}}{{{a^2}}}} \right)} }}} \right] \times \left[ {\dfrac{1}{a}} \right] = \left[ {\dfrac{a}{{\sqrt {{a^2} - {x^2}} }}} \right] \times \left[ {\dfrac{1}{a}} \right] \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {{a^2} - {x^2}} }} \\\

Therefore, the differentiation of the given function [tan⁡−1(xa2−x2)]\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right], with respect to xx is 1a2−x2\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}.

Note- In this problem we have especially substituted x=asin⁡θx = a\sin \theta because due to this substitution, the function inside the tangent inverse will be converted into tangent so that these two will cancel out with each other and we will be left with the angle which is in a much simplified form.