Question

How do you find the derivative of `sin^-1(2x+1)`?

Answer 1

The answer is `2/sqrt(1-(2x+1)^2)`

Explanation:

For this equation you would use the [chain rule] (https://socratic.org/calculus/basic-differentiation-rules/chain-rule) so you take the derivative of the outside:
`(sin^-1)`
times the derivative of the inside:
`(2x + 1)`

So the derivative of `sin^-1` otherwise known as `arcsin` is `1/sqrt(1-x^2)`
Differentiating Inverse Trigonometric Functions

but in this case `(2x-1)` is acting as `x` so it's
`1/sqrt(1-(2x-1)^2)`

Next the derivative of `2x-1` is `2`

So the answer becomes outside times inside
Which is

`2/sqrt(1-(2x-1)^2)`

Here are the derivatives of inverse functions

Answer 2

`color(green)(dy/dx=2/sqrt(1-(2x+1)^2))`

Explanation:

Given:`" Determine "d/dx[sin^(-1)(2x+1)]`

`color(blue)("Preamble")`

Note that `sin^(-1)` has a particular meaning which has the alternative presentation of Arcsin. It is not connected to the form example of `sin^2` which is sin squared

So `sin^(-1)(2x+1) -> arcsin(2x+1)`

Note that `sin^(-1)(2x+1)` is another way of writing an angle and the sin of which gives the value `2x+1`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Answering the question")`

Set `theta =sin^(-1)(2x+1)`

Note that `sin(theta) = ("Opposite")/("Hypotenus")`

Set the value of the Hypotenuse to 1 giving:

`sin(theta) = ("Opposite")/("Hypotenuse") =(2x+1)/1`

Using `sin(theta)=2x+1` and implicitly differentiating

`cos(theta)xxdy/dx =2`

`=>dy/dx=2/cos(theta)`.......................Equation(1)

But `"Hypotenuse "xxcos(theta) =" Adjacent "=x`

as the hypotenuse is 1 we have `cos(theta)=x`

So equation(1) becomes:

`dy/dx=2/x" .......................Equation"(1_a)`

But by Pythagoras `x^2=1^2-(2x+1)^2" "=>" "x=sqrt(1-(2x+1)^2)`

Thus `"Equation"(1_a)` becomes

`color(blue)(dy/dx=2/sqrt(1-(2x+1)^2))`