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

Answer

Verified

364.8k+ views

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 \[y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]\]

Put $x = a\sin \theta $ in the above function, we get

\[ \Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2}\left[ {1 - {{\left( {\sin \theta } \right)}^2}} \right]} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}} \right)} \right]\]

As, we know that ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \Rightarrow 1 - {\left( {\sin \theta } \right)^2} = {\left( {\cos \theta } \right)^2}$

Using this above identity to simplify the function whose differentiation needs to be carried out

\[

\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 $x = a\sin \theta \Rightarrow \sin \theta = \dfrac{x}{a} \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(2)}}$

Substituting the value obtained from equation (2) in equation (1), we get

\[ \Rightarrow y = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(3)}}\]

Also we know that $\dfrac{{d\left[ {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right]}}{{dx}} = \left[ {\dfrac{1}{{\sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} }}} \right] \times \left[ {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}} \right]$

Differentiating both sides of equation (3) with respect to $x$, we get

\[

\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 \[\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]\], with respect to $x$ is \[\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}\].

Note- In this problem we have especially substituted $x = 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.

Let the function be \[y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]\]

Put $x = a\sin \theta $ in the above function, we get

\[ \Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2}\left[ {1 - {{\left( {\sin \theta } \right)}^2}} \right]} }}} \right)} \right] = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}} \right)} \right]\]

As, we know that ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1 \Rightarrow 1 - {\left( {\sin \theta } \right)^2} = {\left( {\cos \theta } \right)^2}$

Using this above identity to simplify the function whose differentiation needs to be carried out

\[

\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 $x = a\sin \theta \Rightarrow \sin \theta = \dfrac{x}{a} \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(2)}}$

Substituting the value obtained from equation (2) in equation (1), we get

\[ \Rightarrow y = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right){\text{ }} \to {\text{(3)}}\]

Also we know that $\dfrac{{d\left[ {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right]}}{{dx}} = \left[ {\dfrac{1}{{\sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} }}} \right] \times \left[ {\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}}} \right]$

Differentiating both sides of equation (3) with respect to $x$, we get

\[

\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 \[\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]\], with respect to $x$ is \[\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}\].

Note- In this problem we have especially substituted $x = 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.

Last updated date: 23rd Sep 2023

â€¢

Total views: 364.8k

â€¢

Views today: 4.64k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE