How do you find the derivative of #x+sqrt(x)#?

1 Answer
Johnson Z.
Mar 3, 2016

#(x+sqrt(x))'=1+1/(2sqrt(x))#

Explanation:

Recall the following:

#1#. #color(red)("Sum Rule")#: #[f(x)+g(x)]'=f'(x)+g'(x)#
#2#. #color(blue)("Power Rule")#: #(x^n)'=color(orange)nx^(color(orange)n-1)#

Finding the Derivative
#1#. Start by rewriting #sqrt(x)# as #x^(1/2)#. Recall that #x# is #x^1#.

#(x+sqrt(x))'#

#=(x^1+x^(1/2))'#

#2#. Using the #color(red)("sum rule")#, determine the derivative of each term in the expression. To do that, use the #color(blue)("power rule")#.

#=color(orange)1x^(color(orange)1-1)+color(orange)(1/2)x^(color(orange)(1/2)-1)#

#3#. Simplify.

#=color(orange)1x^0+color(orange)(1/2)x^(-1/2)#

#=color(orange)1(1)+color(orange)(1/2)(1/x)^(1/2)#

#=1+color(orange)(1/2)(1/sqrt(x))#

#color(green)(=1+1/(2sqrt(x)))#

#:.#, the derivative is #1+1/(2sqrt(x))#.

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