What is the derivative of #sqrt x^2#?

2 Answers
Jan 11, 2016

Answer:

#x/sqrt(x^2) #

Explanation:

first rewrite # sqrt(x^2) = (x^2)^(1/2) #

now differentiate using the chain rule :

# 1/2(x^2)^(-1/2).d/dx(x^2) #

#= 1/2(x^2)^(-1/2) .(2x) =x( x^2)^(-1/2) =x/(x^2)^(1/2) = x/sqrt(x^2 #

It is usual to give answer in the same form as question.

Jan 11, 2016

Answer:

Assuming the question is asking #(sqrt(x))^2# or #sqrt((x^2))#, which is then simplified to #x#, the answer is #1#.

Explanation:

#(sqrt(x))^2# can be rewritten as #(x^(1/2))^(2)#. Multiplying the exponents gives #x^1#, which is just #x#.

The definition of #d/dx(x) = 1# is typically memorized, though you can always use the power rule to get the same answer:

#d/dx(x) = d/dx(x^1) = (1)x^(1-1) = (1)(x^0) = (1)(1) = 1#

If the problem were instead asking #d/dx(sqrt(x))#, with the #2# indicating that it's the second root, then the same power rule approach would be taken:

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