Question #36b8c

2 Answers
Oct 15, 2014

By multiplying out,

H(x)=(x-sqrt{x})(x+sqrt{x})=x^2-x

By Power Rule,

H'(x)=2x-1.

I hope that this was helpful.

Nov 19, 2014

If you notice that H(x) is the difference of perfect squares then the problem is much easier.

If you do not then you can use the Product Rule .

H'(x)=uv'+vu'

H(x)=uv=(x-sqrt(x))(x+sqrt(x))=(x-x^(1/2))(x+x^(1/2))

H'(x)=(x-x^(1/2))(1+1/2x^(-1/2))+(x+x^(1/2))(1-1/2x^(-1/2))

H'(x)=(x-x^(1/2))(1+1/(2x^(1/2)))+(x+x^(1/2))(1-1/(2x^(1/2)))

H'(x)=x+x/(2x^(1/2))-x^(1/2)-x^(1/2)/(2x^(1/2))+x-x/(2x^(1/2))+x^(1/2)-x^(1/2)/(2x^(1/2))

H'(x)=x+x/(2x^(1/2))-x^(1/2)-1/2+x-x/(2x^(1/2))+x^(1/2)-1/2

H'(x)=x+x/(2x^(1/2))-x^(1/2)+x-x/(2x^(1/2))+x^(1/2)-1

H'(x)=x+x/(2x^(1/2))+x-x/(2x^(1/2))-1

H'(x)=x+x-1

H'(x)=2x-1