Question #c1a8e

1 Answer
Oct 4, 2014

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)#