Question

find the derivative of sin^-1(2x+3)

Answer 1

what i did was change Sin^-1(2x+3) to Sin^-1(x+3/1/2) so its in the form of x/a

Answer 2

\[y = \sin^{-1}(2x+3) \iff \sin(y) = 2x+3 \iff \cos(y)\frac{dy}{dx} = 2 \]
\[\iff \frac{dy}{dx} = \frac{2}{\cos(\sin^{-1}(2x+3))} = \frac{2}{\sqrt{1-(2x+3)^2}}\]

i think <.< never liked calculus.

Answer 3

im probably doing it the long way. most likely theres some short formula out there.

Answer 4

well there this formula Sin^-1(x/a) = 1/sqrt(a^2-x^2)

Answer 5

use the chain rule then. let:
\[\frac{d}{dx}\sin^{-1}\left(\frac{f(x)}{a}\right) = \frac{1}{\sqrt{a^2-f(x)^2}}*f'(x)\]

Answer 6

f(x) = 2x+3, a = 1

Answer 7

my answer ended up being :

2/sqrt(1-4(x+3)^2)

Answer 8

but i think the x has to be isolated doesn't it?

Answer 9

using the chain rule, you want to keep the function together.

Answer 10

whatever you define the function to be. Since im defining it as f(x) = 2x+3, i dont want to separate it.

Answer 11

ok i see thanks