a. Difference Quotient
Identify f(x) and f(x+h)
#f(x)=sqrt(3x)#
#f(x+h)=sqrt(3(x+h))#
Make substitutions and rationalize the numerator
#f'(x)=lim_(h->0)(sqrt(3(x+h))-sqrt(3x))/h*(sqrt(3(x+h))+sqrt(3x))/(sqrt(3(x+h))+sqrt(3x))#
Remember difference of perfect squares
#f'(x)=lim_(h->0)(3(x+h)-(3x))/(h*(sqrt(3(x+h))+sqrt(3x)))#
Distribute the #3#
#f'(x)=lim_(h->0)(3x+3h-3x)/(h*(sqrt(3(x+h))+sqrt(3x)))#
Simplify
#f'(x)=lim_(h->0)(3h)/(h*(sqrt(3(x+h))+sqrt(3x)))#
Cancel the factor #h#
#f'(x)=lim_(h->0)(3)/(sqrt(3(x+h))+sqrt(3x))#
Now you can substitute in 0 for #h#
#(3)/(sqrt(3(x+0))+sqrt(3x))=#
Simplify and manipulate the square roots
#(3)/(sqrt(3(x))+sqrt(3x))=(3)/(sqrt(3x)+sqrt(3x))=3/(2sqrt(3x))*sqrt(3)/sqrt(3)=(3sqrt(3))/(6sqrt(x))=sqrt(3)/(2sqrt(x))=sqrt(3)/(2sqrt(x))*sqrt(x)/sqrt(x)=sqrt(3x)/(2x)#
b. Product Rule
#f(x)=u*v=sqrt(3)*sqrt(x)=sqrt(3)*x^(1/2)#
#u'=0#
#v'=(1/2)x^(1/2-1)=(1/2)x^(1/2-2/2)=(1/2)x^(-1/2)=1/(2sqrt(x))#
Apply the product rule by making the appropriate substitutions
#f'(x)=u*v'+v*u'=sqrt(3) * 1/(2sqrt(x))+sqrt(x) * 0=sqrt(3)/(2sqrt(x))=sqrt(3)/(2sqrt(x))*sqrt(x)/sqrt(x)=sqrt(3x)/(2x)#
c. Factor out the constant of #sqrt(3)#
#f(x)=sqrt(3)*x^(1/2)#
#f'(x)=sqrt(3) * d/dx (x^(1/2))=sqrt(3) * (1/2)x^(1/2-1)=sqrt(3) * (1/2)x^(1/2-2/2)=sqrt(3) * (1/2)x^(-1/2)=sqrt(3) * (1/(2x^(1/2)))=sqrt(3)/(2x^(1/2))=sqrt(3)/(2sqrt(x))=sqrt(3)/(2sqrt(x))*sqrt(x)/sqrt(x)=sqrt(3x)/(2x)#
