Question

How do you find the derivative of \[{\sin ^{ - 1}}\left( {2x + 1} \right)\]?

Answer 1

Hint: In order to solve this question first, we assume the function. Then we differentiate that function with respect to \[x\]. First, we find the derivative of the inverse trigonometry function, and then we use the chain rule to differentiate the value. If power is on the function then at first we differentiate that power and continue with the chain rule.

Complete answer:
To find,
The derivative of the function \[{\sin ^{ - 1}}\left( {2x + 1} \right)\].
Let, \[y\] be the function of \[x\]
\[y = {\sin ^{ - 1}}\left( {2x + 1} \right)\]
Now on differentiating with respect to \[x\] both sides.
\[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{{\sin }^{ - 1}}\left( {2x + 1} \right)} \right)}}{{dx}}\]
Now, apply derivative e to the second term in the multiple and using identity –
\[\dfrac{{d\left( {{{\sin }^{ - 1}}f(x)} \right)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {f(x)} \right)}^2}} }}.{f'}(x)\]
By putting this formula in the obtained equation
\[\dfrac{{d\left( {{{\sin }^{ - 1}}\left( {2x + 1} \right)} \right)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {2x + 1} \right)}^2}} }}.\dfrac{{d\left( {2x + 1} \right)}}{{dx}}\]
Now applying the derivative again.
\[\dfrac{{d\left( {{{\sin }^{ - 1}}\left( {2x + 1} \right)} \right)}}{{dx}} = \dfrac{2}{{\sqrt {1 - {{\left( {2x + 1} \right)}^2}} }}.\dfrac{{d\left( x \right)}}{{dx}}\]
On simplifying
\[\dfrac{{d\left( {{{\sin }^{ - 1}}\left( {2x + 1} \right)} \right)}}{{dx}} = \dfrac{2}{{\sqrt { - 4{x^2} - 4x} }}\]
Final answer:
the derivative of \[{\sin ^{ - 1}}\left( {2x + 1} \right)\] is
\[\dfrac{{d\left( {{{\sin }^{ - 1}}\left( {2x + 1} \right)} \right)}}{{dx}} = \dfrac{2}{{\sqrt { - 4{x^2} - 4x} }}\]

Note: To solve these types of questions we must know all the derivatives of the standard trigonometry and inverse trigonometric function. We must know how we apply the chain rule to the function. Differentiation can represent the rate of change of the function. (example:- Derivative of the equation of a circle on a fixed point then that will give a tangent at that point.) integration represents the sum of the function over the range. (example – if we integrate a function from range a to b then that will give the area of that curve on the x-axis.) Both of these are the inverse of each other and if we differentiate an integration then that will give the same equation.