# How do you find the derivative for f(x) = (x^1.7 + 2) / (x^2.8 + 1)?

Oct 15, 2015

$f ' \left(x\right) = \frac{- 11 {x}^{3.5} + 17 {x}^{0.7} - 56 {x}^{1.8}}{10 {\left({x}^{2.8} + 1\right)}^{2}}$

#### Explanation:

$f \left(x\right) = \frac{{x}^{\frac{17}{10}} + 2}{{x}^{\frac{14}{5}} + 1}$

$f ' \left(x\right) = \frac{\frac{17}{10} {x}^{\frac{17}{10} - 1} \cdot \left({x}^{\frac{14}{5}} + 1\right) - \left({x}^{\frac{17}{10}} + 2\right) \cdot \frac{14}{5} {x}^{\frac{14}{5} - 1}}{{x}^{\frac{14}{5}} + 1} ^ 2$

$f ' \left(x\right) = \frac{\frac{17}{10} {x}^{\frac{7}{10}} \cdot \left({x}^{\frac{14}{5}} + 1\right) - \left({x}^{\frac{17}{10}} + 2\right) \cdot \frac{14}{5} {x}^{\frac{9}{5}}}{{x}^{\frac{14}{5}} + 1} ^ 2$

$f ' \left(x\right) = \frac{\frac{17}{10} \left({x}^{\frac{35}{10}} + {x}^{\frac{7}{10}}\right) - \frac{14}{5} \left({x}^{\frac{35}{10}} + 2 {x}^{\frac{9}{5}}\right)}{{x}^{\frac{14}{5}} + 1} ^ 2$

$f ' \left(x\right) = \frac{\frac{17}{10} {x}^{\frac{35}{10}} + \frac{17}{10} {x}^{\frac{7}{10}} - \frac{14}{5} {x}^{\frac{35}{10}} - \frac{28}{5} {x}^{\frac{9}{5}}}{{x}^{\frac{14}{5}} + 1} ^ 2$

$f ' \left(x\right) = \frac{17 {x}^{\frac{35}{10}} + 17 {x}^{\frac{7}{10}} - 28 {x}^{\frac{35}{10}} - 56 {x}^{\frac{9}{5}}}{10 {\left({x}^{\frac{14}{5}} + 1\right)}^{2}}$

$f ' \left(x\right) = \frac{- 11 {x}^{\frac{7}{2}} + 17 {x}^{\frac{7}{10}} - 56 {x}^{\frac{9}{5}}}{10 {\left({x}^{\frac{14}{5}} + 1\right)}^{2}}$

$f ' \left(x\right) = \frac{- 11 {x}^{3.5} + 17 {x}^{0.7} - 56 {x}^{1.8}}{10 {\left({x}^{2.8} + 1\right)}^{2}}$