How do you find the derivative of y=arcsin(2x+1)?
1 Answer
Apr 24, 2016
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)
"-----------------------------------------------------------------" 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
"-------------------------------------------------------------------"
Substitute these values into the derivative
rArr dy/dx = 1/(sqrt(1-(2x+1)^2)) . 2 = 2/(sqrt(1 - (2x+1)^2