How do you find derivative of Y=1/ √ 1-X from the First Principles?

1 Answer
Cesareo R.
Apr 11, 2017

See below.

Explanation:

#dy/dx=lim_(h->0)(y(x+h)-y(x))/h = lim_(h->0)(1/sqrt(1-x-h)-1/sqrt(1-x))/h#

but

#(1/sqrt(1-x-h)-1/sqrt(1-x))/h=1/h(1/sqrt(1-x-h)-1/sqrt(1-x))=#

#=1/h(1/sqrt(1-x-h)-1/sqrt(1-x))((1/sqrt(1-x-h)+1/sqrt(1-x))/(1/sqrt(1-x-h)+1/sqrt(1-x)))=#
#=1/h((1/(1-x-h)-1/(1-x))/(1/sqrt(1-x-h)+1/sqrt(1-x)))=#
#=1/h(1-x-1+x+h)/((1-x-h)(1-x))//(1/sqrt(1-x-h)+1/sqrt(1-x))=#
#=1/((1-x-h)(1-x))((sqrt(1-x-h)sqrt(1-x))/(sqrt(1-x-h)+sqrt(1-x)))#

now

#dy/dx=lim_(h->0)1/((1-x-h)(1-x))((sqrt(1-x-h)sqrt(1-x))/(sqrt(1-x-h)+sqrt(1-x)))=1/(1-x)^2(1-x)/(2sqrt(1-x)) = 1/2(1-x)^(-3/2)#

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