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Question

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

How do you differentiate y=sin1(2x+1)y = {\sin ^{ - 1}}(2x + 1) ?

Explanation

Solution

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 sin1x{\sin ^{ - 1}}x times the derivative of the inside (2x+1)(2x + 1)
y=sin1(2x+1)y = {\sin ^{ - 1}}(2x + 1)
As we know,
The derivative of sin1x{\sin ^{ - 1}}x is 11x2\dfrac{1}{{\sqrt {1 - {x^2}} }}
In this case (2x+1)(2x + 1) is acting as xx so it is 11(2x+1)2\dfrac{1}{{\sqrt {1 - {{(2x + 1)}^2}} }}
Next we will derivate (2x+1)(2x + 1)
Which will be 2 as,
d(2x+1)dx=d2xdx+ddx(1)\dfrac{{d(2x + 1)}}{{dx}} = \dfrac{{d2x}}{{dx}} + \dfrac{{d}}{{dx}}(1)
The derivative of constant term is zero
And the derivative of 2x2x is 22
Hence, The derivative of y=sin1(2x+1)y = {\sin ^{ - 1}}(2x + 1) is 1×21+(2x+1)2\dfrac{1 \times 2}{{\sqrt {1 + {{(2x + 1)}^2}} }}

Note:
Sometimes students get confused between the derivative of sin(x)\sin(x) and sin1(x){\sin ^{ - 1}}(x). Students should remember that chain rule is used only for composite functions.