What is the derivative f'of the function f? f(x)=1+x^(1/2)

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Answer 1 · 2018-02-03T23:04:30

#f'(x)=sqrtx/(2x)#

.

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

#f'(x)=0+1/2x^(1/2-1)=x^(-1/2)/2=1/(2x^(1/2))=1/(2sqrtx)=sqrtx/(2x)#

Answer 2 · 2018-02-03T23:08:16

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

Since the derivative of a sum is the sum of the derivatives we have #f'(x)=(1)'+(x^(1/2))'#

Since the derivative of a constant is #0# we can say #f'(x)=(x^(1/2))'#

Then by the power rule

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