Question

How do you differentiate $y = {\sin ^{ - 1}}(2x + 1)$ ?

Answer 1

Hint: To differentiate this we will apply a chain rule. In chain rule we first differentiate the whole term as a single function and let another function be constant . After that we differentiate another function which we had assumed as constant and multiply it with the previous result.
Complete step by step answer:
For this equation you would use the chain rule.
Taking the derivative of ${\sin ^{ - 1}}x$ times the derivative of the inside $(2x + 1)$
$y = {\sin ^{ - 1}}(2x + 1)$
As we know,
The derivative of ${\sin ^{ - 1}}x$ is $\dfrac{1}{{\sqrt {1 - {x^2}} }}$
In this case $(2x + 1)$ is acting as $x$ so it is $\dfrac{1}{{\sqrt {1 - {{(2x + 1)}^2}} }}$
Next we will derivate $(2x + 1)$
Which will be 2 as,
 $\dfrac{{d(2x + 1)}}{{dx}} = \dfrac{{d2x}}{{dx}} + \dfrac{{d}}{{dx}}(1)$
The derivative of constant term is zero
And the derivative of $2x$ is $2$
Hence, The derivative of $y = {\sin ^{ - 1}}(2x + 1)$ is $\dfrac{1 \times 2}{{\sqrt {1 + {{(2x + 1)}^2}} }}$

Note:
Sometimes students get confused between the derivative of $\sin(x)$ and ${\sin ^{ - 1}}(x)$. Students should remember that chain rule is used only for composite functions.