How do you find the derivative of #sqrt(1-2x)#?
2 Answers
Apr 28, 2016
#dy/dx=(2x)/(2sqrt(1-2x)#
Explanation:
Given -
#y=sqrt(1-2x)#
#dy/dx=(2x)/(2sqrt(1-2x)#
Apr 28, 2016
Explanation:
differentiate using the
#color(blue)" chain rule "#
# d/dx [f(g(x)) ] = f'(g(x)).g'(x)#
#"-----------------------------------------------"# rewrite
#sqrt(1-2x) = (1-2x)^(1/2) # f(g(x)) =
#(1-2x)^(1/2) rArr f'(g(x)) = 1/2(1-2x)^(-1/2) # and g(x) = 1-2x → g'(x) = -2
#rArr d/dx(sqrt(1-2x))=1/2(1-2x)^(-1/2).(-2)#
#= (-1)/(1-2x)^(1/2) = (-1)/(sqrt(1-2x))#