How do you differentiate #y=(x^2+4x+3)/sqrtx#?

1 Answer
Aug 30, 2017

#dy/dx=(3x^2+4x-3)/(2x^(3/2))#

Explanation:

What I'd prefer is to do is split up the fraction into its three parts by dividing each term by #sqrtx=x^(1/2)#:

#y=x^2/x^(1/2)+(4x)/x^(1/2)+3/x^(1/2)#

#y=x^(3/2)+4x^(1/2)+3x^(-1/2)#

Now instead of having to use the quotient rule, we can simply differentiate term-by-term using the power rule:

#dy/dx=3/2x^(1/2)+4*1/2x^(-1/2)+3*-1/2x^(-3/2)#

#dy/dx=(3x^(1/2))/2+2/x^(1/2)-3/(2x^(3/2))#

Finding a common denominator of #2x^(3/2)#:

#dy/dx=(3x^2+4x-3)/(2x^(3/2))#