Question

How do you find the derivative of $ \arcsin (2x) $?

Answer 1

Hint: To differentiate the above equation we need to know that the above equation is an inverse trigonometric sine function of 2x. This function can be differentiated by using the chain rule with respect to the variable x.

Complete step by step answer:
 $ \arcsin (2x) $ is an inverse trigonometric function which can also be written as \[{\sin ^{ - 1}}x\]. -1 Here is just a way of showing that it is inverse of sin\[x\].Inverse sine does the opposite of sin. Sin function gives the angle which is calculated by dividing the opposite side and hypotenuse in a right-angle triangle, but the inverse of it gives the measure of an angle.
The above equation is a composite function which means it contains a function inside another function.
Suppose f(x) and g(x) are two different functions. So, the composite of this will be f(g(x)).
Similarly,2x is inside arcsin function
First step is to differentiate by applying chain rule. Chain rule can be written as:
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
So, by differentiating using chain rule we get,
Let \[2x = u\]
\[\dfrac{d}{{dx}}(\arcsin (2x)) = \dfrac{d}{{du}}(\arcsin (u)).\dfrac{d}{{dx}}2x\]
Now, differentiate $ \arcsin (2x) $ which is a common derivative
\[\dfrac{d}{{du}}(\arcsin (u)) = \dfrac{1}{{\sqrt {1 - {u^2}} }}\]
And differentiate the next term,
\[\dfrac{d}{{dx}}(2x) = 2\]
Therefore, substituting the differentiated value we get,
      \[ = \dfrac{1}{{\sqrt {1 - {u^2}} }} \cdot 2\]
 Substitute u=2\[x\]
\[ = \dfrac{1}{{\sqrt {1 - {{(2x)}^2}} }} \cdot 2\]
Further multiplying it we get,
\[ = \dfrac{2}{{\sqrt {1 - 4{x^2}} }}\]
Therefore, we get the above expression as the derivative of $ \arcsin (2x) $ .

Note:
In the above question, an alternative method of derivation is by following the general rule of this type of trigonometric function where derivative of the inverse of the sine function is \[\dfrac{{u'}}{{\sqrt {1 - {u^2}} }}\]
Further by substituting the value of u=2\[x\] and \[u'\]=2 we get the same solution as above.