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

#(-1)/sqrt(1-2x)#

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))#