# If f(x) = [5x^2+10x+26]/sqrt(x) how do you find f '(x)?

Apr 6, 2015

You could use the quotient rule, but I think it's easier to first rewrite the expression algebraically.

$f \left(x\right) = \frac{5 {x}^{2} + 10 x + 26}{\sqrt{x}} = \frac{5 {x}^{2}}{x} ^ \left(\frac{1}{2}\right) + \frac{10 x}{x} ^ \left(\frac{1}{2}\right) + \frac{26}{x} ^ \left(\frac{1}{2}\right)$

$f \left(x\right) = 5 {x}^{\frac{3}{2}} + 10 {x}^{\frac{1}{2}} + 26 {x}^{- \frac{1}{2}}$

So the derivative is:

$f ' \left(x\right) = \frac{3}{2} 5 {x}^{\frac{1}{2}} + \frac{1}{2} 10 {x}^{- \frac{1}{2}} + \frac{- 1}{2} 26 {x}^{- \frac{3}{2}}$

$f ' \left(x\right) = \frac{15}{2} {x}^{\frac{1}{2}} + 5 {x}^{- \frac{1}{2}} - 13 {x}^{\frac{- 3}{2}}$

Rewrite this algebraically (get a common denominator) if you wish.

Apr 6, 2015

I would use the Quotient Rule: