# How do you find the derivative of  f(t)= (t^4 +4t^2 -2)^(4/7) using the chain rule?

Jan 29, 2016

here is how you do it friend,

#### Explanation:

here,
$f \left(t\right) = {\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{4}{7}}$

so,

$\frac{d}{\mathrm{dt}} \left(f \left(t\right)\right)$

$= \frac{d}{\mathrm{dt}} {\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{4}{7}}$

$= \frac{4}{7} {\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{4}{7} - 1} \cdot \frac{d}{\mathrm{dt}} \left({t}^{4} + 4 {t}^{2} - 2\right)$

$= \frac{4}{7} {\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{- 3}{7}} \left[\frac{d}{\mathrm{dt}} \left({t}^{4}\right) + 4 \frac{d}{\mathrm{dt}} \left({t}^{2}\right) - \frac{d}{\mathrm{dt}} \left(2\right)\right]$

$= \frac{4}{7} \frac{4 {t}^{3} + 8 t - 0}{{\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{3}{7}}}$

$= \frac{16 {t}^{3} + 32 t}{7 {\left({t}^{4} + 4 {t}^{2} - 2\right)}^{\frac{3}{7}}}$