What is the derivative of (4x^8-sqrt(x))/(8x^4)?

Aug 1, 2015

${y}^{'} = \frac{1}{16} \cdot \left(32 {x}^{\frac{15}{2}} + 7\right) \cdot {x}^{- \frac{9}{2}}$

Explanation:

Start by rewriting your function like this

$y = \frac{4 {x}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}}{8 \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{4}}}}} - {x}^{\frac{1}{2}} / \left(8 {x}^{4}\right)$

$y = \frac{1}{2} {x}^{4} - \frac{1}{8} \cdot {x}^{- \frac{7}{2}}$

Now you can use the power rule to differentiate $y$

$y = \frac{1}{2} \left[\frac{d}{\mathrm{dx}} \left({x}^{4}\right)\right] - \frac{1}{8} \frac{d}{\mathrm{dx}} \left({x}^{- \frac{7}{2}}\right)$

$y = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} {x}^{3} - \frac{1}{8} \cdot \left(- \frac{7}{2}\right) \cdot {x}^{- \frac{9}{2}}$

$y = 2 {x}^{3} + \frac{7}{16} {x}^{- \frac{9}{2}}$

This can be rewritten as

${y}^{'} = \frac{{x}^{- \frac{9}{2}} \cdot \left(16 \cdot 2 {x}^{\frac{15}{2}} + 7\right)}{16} = \textcolor{g r e e n}{\frac{1}{16} \cdot \left(32 {x}^{\frac{15}{2}} + 7\right) \cdot {x}^{- \frac{9}{2}}}$