Answer
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Hint: Here, we will take that \[h = {\csc ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right)\] and \[g = \sqrt {1 - {x^2}} \]. Then we will use that when \[h\] is differentiated with respect to \[g\], we have to calculate the value of \[\dfrac{{dh}}{{dg}}\]. After differentiating \[h\] with respect to \[x\] and \[g\] with respect to \[x\], we will divide them to find the required value.
Complete step-by-step answer:
Let us assume that \[h = {\csc ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right)\] and \[g = \sqrt {1 - {x^2}} \].
We know that when \[h\] is differentiated with respect to \[g\], we have to calculate the value of \[\dfrac{{dh}}{{dg}}\].
Differentiating the equation \[h\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\csc }^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right)} \right)\]
Using the property, \[{\csc ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)\] in the above equation, we get
\[
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left( {{{\sin }^{ - 1}}\left( {2{x^2} - 1} \right)} \right) \\
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left( {\dfrac{\pi }{2} - {{\cos }^{ - 1}}\left( {2{x^2} - 1} \right)} \right) \\
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left\{
\dfrac{\pi }{2} - 2{\cos ^{ - 1}}x{\text{ , if 0 < x < 1}} \\
\Rightarrow - \dfrac{{3\pi }}{2} + 2{\cos ^{ - 1}}x{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dx}} = \left\{
\dfrac{d}{{dx}}\left( {\dfrac{\pi }{2} - 2{{\cos }^{ - 1}}x} \right){\text{ , if 0 < x < 1}} \\
\dfrac{d}{{dx}}\left( { - \dfrac{{3\pi }}{2} + 2{{\cos }^{ - 1}}x} \right){\text{ , if - 1 < x < 0}} \\
\right. \\
\]
Using the value, \[\dfrac{d}{{dx}}{\cos ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}\] in the above equation, we get
\[ \Rightarrow \dfrac{{dh}}{{dx}} = \left\{
\dfrac{2}{{\sqrt {1 - {x^2}} }}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{\sqrt {1 - {x^2}} }}{\text{ , if - 1 < x < 0}} \\
\right.{\text{ ......eq.(1)}}\]
Differentiating the equation \[g\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right)\]
Using the property, \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\] in the above equation, we get
\[ \Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{{ - x}}{{\sqrt {1 - {x^2}} }}{\text{ ......eq.(2)}}\]
Dividing the equation (1) by equation (2), we get
\[
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
\dfrac{2}{{\sqrt {1 - {x^2}} }} \times \dfrac{{\sqrt {1 - {x^2}} }}{{ - x}}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{\sqrt {1 - {x^2}} }} \times \dfrac{{\sqrt {1 - {x^2}} }}{{ - x}}{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
\dfrac{2}{{ - x}}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{ - x}}{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
- \dfrac{2}{x}{\text{ , if 0 < x < 1}} \\
\dfrac{2}{x}{\text{ , if - 1 < x < 0}} \\
\right. \\
\]
Since we know that \[0 < \dfrac{1}{2} < 1\], so it lies on the upper parts of the above equation.
Taking \[x = \dfrac{1}{2}\] in the above equation, we get
\[
\Rightarrow {\left. {\dfrac{{dh}}{{dg}}} \right|_{x = \dfrac{1}{2}}} = - \dfrac{2}{{\dfrac{1}{2}}} \\
\Rightarrow {\left. {\dfrac{{dh}}{{dg}}} \right|_{x = \dfrac{1}{2}}} = - 4 \\
\]
Note: You should be familiar with the basic properties of differentiation and trigonometric functions, like \[{\csc ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)\] and \[\dfrac{d}{{dx}}{\csc ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}\]. Students get confused to find the derivative and end up computing with respect to \[x\], which is wrong. A function can only be differentiated with respect to another function if and only if both the functions are dependent on the same variable. The key point is to use the differentiation properly to find the final answer.
Complete step-by-step answer:
Let us assume that \[h = {\csc ^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right)\] and \[g = \sqrt {1 - {x^2}} \].
We know that when \[h\] is differentiated with respect to \[g\], we have to calculate the value of \[\dfrac{{dh}}{{dg}}\].
Differentiating the equation \[h\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\csc }^{ - 1}}\left( {\dfrac{1}{{2{x^2} - 1}}} \right)} \right)\]
Using the property, \[{\csc ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)\] in the above equation, we get
\[
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left( {{{\sin }^{ - 1}}\left( {2{x^2} - 1} \right)} \right) \\
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left( {\dfrac{\pi }{2} - {{\cos }^{ - 1}}\left( {2{x^2} - 1} \right)} \right) \\
\Rightarrow \dfrac{{dh}}{{dx}} = \dfrac{d}{x}\left\{
\dfrac{\pi }{2} - 2{\cos ^{ - 1}}x{\text{ , if 0 < x < 1}} \\
\Rightarrow - \dfrac{{3\pi }}{2} + 2{\cos ^{ - 1}}x{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dx}} = \left\{
\dfrac{d}{{dx}}\left( {\dfrac{\pi }{2} - 2{{\cos }^{ - 1}}x} \right){\text{ , if 0 < x < 1}} \\
\dfrac{d}{{dx}}\left( { - \dfrac{{3\pi }}{2} + 2{{\cos }^{ - 1}}x} \right){\text{ , if - 1 < x < 0}} \\
\right. \\
\]
Using the value, \[\dfrac{d}{{dx}}{\cos ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}\] in the above equation, we get
\[ \Rightarrow \dfrac{{dh}}{{dx}} = \left\{
\dfrac{2}{{\sqrt {1 - {x^2}} }}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{\sqrt {1 - {x^2}} }}{\text{ , if - 1 < x < 0}} \\
\right.{\text{ ......eq.(1)}}\]
Differentiating the equation \[g\] with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right)\]
Using the property, \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\] in the above equation, we get
\[ \Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{{ - x}}{{\sqrt {1 - {x^2}} }}{\text{ ......eq.(2)}}\]
Dividing the equation (1) by equation (2), we get
\[
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
\dfrac{2}{{\sqrt {1 - {x^2}} }} \times \dfrac{{\sqrt {1 - {x^2}} }}{{ - x}}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{\sqrt {1 - {x^2}} }} \times \dfrac{{\sqrt {1 - {x^2}} }}{{ - x}}{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
\dfrac{2}{{ - x}}{\text{ , if 0 < x < 1}} \\
\dfrac{{ - 2}}{{ - x}}{\text{ , if - 1 < x < 0}} \\
\right. \\
\Rightarrow \dfrac{{dh}}{{dg}} = \left\{
- \dfrac{2}{x}{\text{ , if 0 < x < 1}} \\
\dfrac{2}{x}{\text{ , if - 1 < x < 0}} \\
\right. \\
\]
Since we know that \[0 < \dfrac{1}{2} < 1\], so it lies on the upper parts of the above equation.
Taking \[x = \dfrac{1}{2}\] in the above equation, we get
\[
\Rightarrow {\left. {\dfrac{{dh}}{{dg}}} \right|_{x = \dfrac{1}{2}}} = - \dfrac{2}{{\dfrac{1}{2}}} \\
\Rightarrow {\left. {\dfrac{{dh}}{{dg}}} \right|_{x = \dfrac{1}{2}}} = - 4 \\
\]
Note: You should be familiar with the basic properties of differentiation and trigonometric functions, like \[{\csc ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{1}{x}} \right)\] and \[\dfrac{d}{{dx}}{\csc ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}\]. Students get confused to find the derivative and end up computing with respect to \[x\], which is wrong. A function can only be differentiated with respect to another function if and only if both the functions are dependent on the same variable. The key point is to use the differentiation properly to find the final answer.