# How do you differentiate y=-3sqrt(7t^3-1)?

Jul 7, 2015

This can be rewritten as:

$y = - 3 \sqrt{\left(7 {t}^{3} - 1\right)} = - 3 {\left(7 {t}^{3} - 1\right)}^{\frac{1}{2}}$.

So to differentiate this need to apply chain rule.

$y ' = - 3 \left(\left(\frac{1}{2}\right) {\left(7 {t}^{3} - 1\right)}^{\left(\frac{1}{2}\right) - 1}\right) \left(7 \cdot 3 {t}^{2}\right)$.

Which comes out to be

$y ' = - 63 {\left(7 {t}^{3} - 1\right)}^{- \frac{1}{2}} \left({t}^{2}\right)$

Cheerio!