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

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Answer 1 · 2017-08-30T17:48:24

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

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 $2x^(3/2)$ :

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

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