How do you find the derivative of y=arcsin(2x+1)?

1 Answer
Apr 24, 2016

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

Explanation:

differentiate using the color(blue)" chain rule "

d/dx [f(g(x)) ] = f'(g(x)) . g'(x)

and the standard derivative D(sin^-1 x) = 1/(sqrt(1 - x^2)
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f(g(x)) = sin^-1 (2x+1) rArr f'(g(x)) = 1/(sqrt(1-(2x+1)^2)
and g(x) = 2x + 1 → g'(x) = 2
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Substitute these values into the derivative

rArr dy/dx = 1/(sqrt(1-(2x+1)^2)) . 2 = 2/(sqrt(1 - (2x+1)^2