How do you differentiate # y=cos^-1(1-2x^2)#?

2 Answers
Mar 6, 2016

Answer:

#dy/dx = 2/sqrt(1-x^2) #

Explanation:

Using the#color(blue)" chain rule "#

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

and the standard derivative #d/dx(cos^-1 x) =( -1)/sqrt(1 - x^2)#

#rArr dy/dx = (-1)/sqrt(1 - (1-2x^2)^2) d/dx(1-2x^2)#

# = (4x)/sqrt(1-1 +4x^2 - 4x^4) =( 4x)/sqrt(4x^2(1-x^2))#

#= (4x)/(2xsqrt(1-x^2))= 2/sqrt(1-x^2)#

Mar 6, 2016

Answer:

#d y=(4x)/(1-2x^2)^2 sin^-1(1-2x^2) d x#

Explanation:

#y=cos u" " y^'=- u^' * sin u#
#y=cos^-1(1-2x^2)" "y=cos(1/(1-2x^2))#
#d y=-((0*(1-2x^2)+4x*1)/(1-2x^2)^2)sin(1/(1-2x^2))*d x#
#d y=(4x)/(1-2x^2)^2 sin^-1(1-2x^2) d x#