Question

How do you differentiate \[y = \arcsin \left( {\dfrac{x}{2}} \right)?\]

Answer 1

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.