Question #36b8c

Question

1107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-15T11:07:17

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.

Answer 2 · 2014-11-19T10:50:34

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#