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

How do you differentiate y=arcsin(x2)?y = \arcsin \left( {\dfrac{x}{2}} \right)?

Explanation

Solution

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)\arcsin (y)” is written as sin1(y)si{n^{ - 1}}(y).
So, We need to apply the differentiation formula of ddysin1(y)\dfrac{d}{{dy}}si{n^{ - 1}}(y).
We know the following formula:
ddysin1(y)=11y2+C\dfrac{d}{{dy}}si{n^{ - 1}}(y) = \dfrac{1}{{\sqrt {1 - {y^2}} }} + C, where CCis an arbitrary constant.
And, we also know that the derivation taken with respect to a variable upon the same variable gives the value of 11, 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:
ddy(m.y)=m+C\dfrac{d}{{dy}}(m.y) = m + C, where CC is an arbitrary constant.

Complete step-by-step solution:
It is given in the question that, y=arcsin(x2)y = \arcsin \left( {\dfrac{x}{2}} \right).
So, according to above formula, we can write the following expression:
y=sin1(x2)\Rightarrow y = si{n^{ - 1}}\left( {\dfrac{x}{2}} \right).
So, we need to do the derivation for sin1(x2)si{n^{ - 1}}\left( {\dfrac{x}{2}} \right) as well as (x2)\left( {\dfrac{x}{2}} \right) with respect to xx.
So, after take the derivation on both sides with respect to xx, we get:
dydx=ddx(sin1(x2))\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:
dydx=ddx(sin1(x2))ddx(x2)\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:
dydx=11(x2)2dxdx(12)+K\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{x}{2}} \right)}^2}} }}\dfrac{{dx}}{{dx}}\left( {\dfrac{1}{2}} \right) + K, where KK is an arbitrary constant.
Now, simplify the squared term under the denominator, we get:
dydx=11(x24)×12+K\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - \left( {\dfrac{{{x^2}}}{4}} \right)} }} \times \dfrac{1}{2} + K, where KK is an arbitrary constant.
Now, by doing further simplification, we get:
dydx=12(4x24)+K\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {\left( {\dfrac{{4 - {x^2}}}{4}} \right)} }} + K, where KK is an arbitrary constant.
Now, by doing further simplification, we get:
dydx=12×4x22+K\dfrac{{dy}}{{dx}} = \dfrac{1}{{2 \times \dfrac{{\sqrt {4 - {x^2}} }}{2}}} + K, where KK is an arbitrary constant.
Now, by doing further simplification, we get:
dydx=14x2+K\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {4 - {x^2}} }} + K, where KK is an arbitrary constant.

\therefore The differentiation of y=arcsin(x2)y = \arcsin \left( {\dfrac{x}{2}} \right) is 14x2+K\dfrac{1}{{\sqrt {4 - {x^2}} }} + K, where KK 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.