What is the derivative of #sqrt(x - 2)#?

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Answer 1 · 2016-06-24T21:24:25

#f^(')=sqrt(x-2)/(2x-4)#

I prefer to use a particular notation.

Let #u=x-2" "# Then #" "(du)/(dx)=1#

Let #" "y=sqrt(x-2)" "=" "sqrt(u)" "=" "u^(1/2)#

#(dy)/(du)=1/2 u^(1/2-1)-> 1/2u^(-1/2)#

But #(dy)/(dx)" "=" "(du)/(dx)xx(dy)/(du)#

#=>(dy)/(du)= 1xx1/(2sqrt(x-2))#

Multiply by 1 in the form of #1=sqrt(x-2)/sqrt(x-2)#

#=>(dy)/(du)= (sqrt(x-2))/(2(x-2))" "=" "sqrt(x-2)/(2x-4)#

Or if you prefer:

#f^(')=sqrt(x-2)/(2x-4)#