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Question: Find the differentiation of \(\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2}...

Find the differentiation of ddx[tan⁡−1(xa2−x2)]\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]
A) x1−x2\dfrac{x}{{\sqrt {1 - {x^2}} }}
B) 1−x2x\dfrac{{\sqrt {1 - {x^2}} }}{x}
C) 11+x2\dfrac{1}{{1 + {x^2}}}
D) 1a2−x2\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}

Explanation

Solution

According to given in the question we have to determine the value of the differential equation which is ddx[tan⁡−1(xa2−x2)]\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right]. So, first of all we have to let that given differentiation is y.
Now, we have to let that x is equal to asin⁡θa\sin \theta so we can simplify the differential expression easily and on substituting the value of x as we have to solve some trigonometric terms that will be obtained in the expression.
Now, to solve the terms in the differential equation obtained we have to use the formula as mentioned below:

Formula used: ⇒(1−sin⁡2θ)=cos⁡2θ...............(A) \Rightarrow (1 - {\sin ^2}\theta ) = {\cos ^2}\theta ...............(A)
Now, we have to obtain the value of θ\theta with the help of x=asin⁡θx = a\sin \theta as we let. Hence,
⇒θ=sin⁡−1xa...............(B)\Rightarrow \theta = {\sin ^{ - 1}}\dfrac{x}{a}...............(B)
Now, to solve the obtained expression we have to use the formula as mentioned below:
Formula used:
⇒ddx(sin⁡−1x)=11−x2.................(C)\Rightarrow \dfrac{d}{{dx}}({\sin ^{ - 1}}x) = \dfrac{1}{{\sqrt {1 - {x^2}} }}.................(C)
Hence, with the help of the formula (B) above, and eliminating the terms in the expression we can obtain the value of differentiation with respect to x.

Complete step-by-step solution:
Step 1: First of all we have to let the given differential equation is y as mentioned in the solution hint. Hence,
⇒y=[tan⁡−1(xa2−x2)].............(1)\Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right].............(1)
Step 2: Now, we have to let that the variable x as given in the differential equation is asin⁡θa\sin \theta as mentioned in the solution hint. Hence, on substituting the value of x as we let in the expression (1),
⇒y=[tan⁡−1(asin⁡θa2−a2sin⁡2θ)].............(2)\Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)} \right].............(2)
Step 3: Now, we have to take a as a common term and then we have to eliminate a in the expression (2) as we obtained in the solution step 2.
⇒y=[tan⁡−1(asin⁡θa1−sin⁡2θ)] ⇒y=[tan⁡−1(sin⁡θ1−sin⁡2θ)]..........(3)  \Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)} \right] \\\ \Rightarrow y = \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \right)} \right]..........(3) \\\
Step 4: Now, to solve the expression (3) as obtained in the solution step 3 we have to use the formula (A) as mentioned in the solution hint. Hence,
⇒y=tan⁡−1[sin⁡θcos⁡2θ] ⇒y=tan⁡−1[sin⁡θcos⁡θ].................(4)  \Rightarrow y = {\tan ^{ - 1}}\left[ {\dfrac{{\sin \theta }}{{\sqrt {{{\cos }^2}\theta } }}} \right] \\\ \Rightarrow y = {\tan ^{ - 1}}\left[ {\dfrac{{\sin \theta }}{{\cos \theta }}} \right].................(4) \\\
Step 5: Now, to simplify the expression (4) obtained in the solution step 4 as we know that
⇒tan⁡θ=sin⁡θcos⁡θ\Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}Hence, on substituting in the equation (4)
⇒y=tan⁡−1(tan⁡θ)\Rightarrow y = {\tan ^{ - 1}}(\tan \theta )
And we all know that tan⁡−1(tan⁡θ){\tan ^{ - 1}}(\tan \theta )= 1 hence, substituting this in the equation obtained just above,
⇒y=θ\Rightarrow y = \theta…………………..(5)
Step 5: Now, we have to substitute the value of θ\theta as mentioned in the solution hint, in the expression (5) as we obtained in the solution step 5. Hence,
⇒y=sin⁡−1xa\Rightarrow y = {\sin ^{ - 1}}\dfrac{x}{a}
On taking differentiation on the both sides of the expression as obtained just above,
⇒ddxy=ddxsin⁡−1xa\Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}{\sin ^{ - 1}}\dfrac{x}{a}
Step 6: Now, to find the differentiation of equation as obtained in the solution step 5 with the help of the formula (C) as given in the solution hint.
=1a×11−x2a2= \dfrac{1}{a} \times \dfrac{1}{{\sqrt {1 - \dfrac{{{x^2}}}{{{a^2}}}} }}
On solving the equation as obtained just above,
=1a×1a2−x2a2 =1a×aa2−x2 =1a2−x2  = \dfrac{1}{a} \times \dfrac{1}{{\sqrt {\dfrac{{{a^2} - {x^2}}}{{{a^2}}}} }} \\\ = \dfrac{1}{a} \times \dfrac{a}{{\sqrt {{a^2} - {x^2}} }} \\\ = \dfrac{1}{{\sqrt {{a^2} - {x^2}} }} \\\
Final solution: Hence, we have obtained the value of the given differentiation of ddx[tan⁡−1(xa2−x2)]\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}} \right)} \right] =1a2−x2 = \dfrac{1}{{\sqrt {{a^2} - {x^2}} }}.

Therefore option (D) is correct.

Note: To solve the differentiation of the given equation with respect to x it is necessary to let that equation to some variable as for the given equation we let as y.
We have to let the variable x as given in the equation have some trigonometric term as asin⁡θa\sin \theta so that we can easily obtain the simplified form of the given equation.