How do you differentiate \[y = \arcsin \left( {\dfrac{x}{2}} \right)?\]
Answer
621.9k+ views
Hint: Here we have to find out the derivatives of the given term. Now, we will apply the derivation formula of inverse function directly. After doing some simplification we get the required answer.
Formula used: In mathematics “\[\arcsin (y)\]” is written as \[si{n^{ - 1}}(y)\].
So, We need to apply the differentiation formula of \[\dfrac{d}{{dy}}si{n^{ - 1}}(y)\].
We know the following formula:
\[\dfrac{d}{{dy}}si{n^{ - 1}}(y) = \dfrac{1}{{\sqrt {1 - {y^2}} }} + C\], where \[C\]is an arbitrary constant.
And, we also know that the derivation taken with respect to a variable upon the same variable gives the value of \[1\], but if the variable has any coefficient associated with the variable then the value of the differentiation will be equal to that constant term only.
So, we can derive the above statement as following:
\[\dfrac{d}{{dy}}(m.y) = m + C\], where \[C\] is an arbitrary constant.
Complete step-by-step solution:
It is given in the question that, \[y = \arcsin \left( {\dfrac{x}{2}} \right)\].
So, according to above formula, we can write the following expression:
\[ \Rightarrow y = si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)\].
So, we need to do the derivation for \[si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)\] as well as \[\left( {\dfrac{x}{2}} \right)\] with respect to \[x\].
So, after take the derivation on both sides with respect to \[x\], we get:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)\].
We can write down the above equation in following way:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)\dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)\].
By applying the above formula, we can rewrite the above equation in following manner:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{x}{2}} \right)}^2}} }}\dfrac{{dx}}{{dx}}\left( {\dfrac{1}{2}} \right) + K\], where \[K\] is an arbitrary constant.
Now, simplify the squared term under the denominator, we get:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - \left( {\dfrac{{{x^2}}}{4}} \right)} }} \times \dfrac{1}{2} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {\left( {\dfrac{{4 - {x^2}}}{4}} \right)} }} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2 \times \dfrac{{\sqrt {4 - {x^2}} }}{2}}} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {4 - {x^2}} }} + K\], where \[K\] is an arbitrary constant.
\[\therefore \] The differentiation of \[y = \arcsin \left( {\dfrac{x}{2}} \right)\] is \[\dfrac{1}{{\sqrt {4 - {x^2}} }} + K\], where \[K\] is an arbitrary constant.
Note: Points to remember:
Derivation will always be taken with respect to the variable that is a dependable variable of any function.
We always need to put an arbitrary constant after the differentiation.
Formula used: In mathematics “\[\arcsin (y)\]” is written as \[si{n^{ - 1}}(y)\].
So, We need to apply the differentiation formula of \[\dfrac{d}{{dy}}si{n^{ - 1}}(y)\].
We know the following formula:
\[\dfrac{d}{{dy}}si{n^{ - 1}}(y) = \dfrac{1}{{\sqrt {1 - {y^2}} }} + C\], where \[C\]is an arbitrary constant.
And, we also know that the derivation taken with respect to a variable upon the same variable gives the value of \[1\], but if the variable has any coefficient associated with the variable then the value of the differentiation will be equal to that constant term only.
So, we can derive the above statement as following:
\[\dfrac{d}{{dy}}(m.y) = m + C\], where \[C\] is an arbitrary constant.
Complete step-by-step solution:
It is given in the question that, \[y = \arcsin \left( {\dfrac{x}{2}} \right)\].
So, according to above formula, we can write the following expression:
\[ \Rightarrow y = si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)\].
So, we need to do the derivation for \[si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)\] as well as \[\left( {\dfrac{x}{2}} \right)\] with respect to \[x\].
So, after take the derivation on both sides with respect to \[x\], we get:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)\].
We can write down the above equation in following way:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {si{n^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)\dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)\].
By applying the above formula, we can rewrite the above equation in following manner:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{x}{2}} \right)}^2}} }}\dfrac{{dx}}{{dx}}\left( {\dfrac{1}{2}} \right) + K\], where \[K\] is an arbitrary constant.
Now, simplify the squared term under the denominator, we get:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - \left( {\dfrac{{{x^2}}}{4}} \right)} }} \times \dfrac{1}{2} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {\left( {\dfrac{{4 - {x^2}}}{4}} \right)} }} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2 \times \dfrac{{\sqrt {4 - {x^2}} }}{2}}} + K\], where \[K\] is an arbitrary constant.
Now, by doing further simplification, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {4 - {x^2}} }} + K\], where \[K\] is an arbitrary constant.
\[\therefore \] The differentiation of \[y = \arcsin \left( {\dfrac{x}{2}} \right)\] is \[\dfrac{1}{{\sqrt {4 - {x^2}} }} + K\], where \[K\] is an arbitrary constant.
Note: Points to remember:
Derivation will always be taken with respect to the variable that is a dependable variable of any function.
We always need to put an arbitrary constant after the differentiation.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

