How do you differentiate #g(t)=(t-sqrtt)/(t^(1/3))#?

Question

Emma Aldridge · Accepted Answer

I would rewrite to avoid the quotient rule.

#g(t) = (t-sqrtt)/t^(1/3) = (t-t^(1/2))/t^(1/3)# # = t/t^(1/3) - t^(1/2)/t^(1/3)# # = t^(1-1/3) - t^(1/2-1/3)# # = t^(2/3) - t^(1/6)# #g'(t) = 2/3t^(-1/3) - 1/6 t^(-5/6)# # = 2/(3root(3)x) - 1/(6 root(6)(x^5))# If you want a single ratio for the derivative, use #root(3)x = root(3)(x^2)# to get a common denominator. # = (4x-1)/(6root(6)(x^5))#

Axel Alvarez · Answer

undefined To differentiate the function $ g(t) = \frac{t - \sqrt{t}}{t^{1/3}} $, we can use the quotient rule of differentiation. The quotient rule states that for functions $ u(t) $ and $ v(t) $, the derivative of $ \frac{u(t)}{v(t)} $ is given by:

$$ \frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} $$

Applying this rule to the given function $ g(t) $, where $ u(t) = t - \sqrt{t} $ and $ v(t) = t^{1/3} $, we have:

$$ g'(t) = \frac{(1 - \frac{1}{2\sqrt{t}})(t^{1/3}) - (t - \sqrt{t})\left(\frac{1}{3}t^{-2/3}\right)}{(t^{1/3})^2} $$

Simplifying further yields:

$$ g'(t) = \frac{(t^{1/3} - \frac{1}{2}t^{-1/6}) - (\frac{1}{3}t^{1/3} - \frac{1}{3}t^{-2/3})}{t^{2/3}} $$

Combining like terms:

$$ g'(t) = \frac{\frac{2}{3}t^{1/3} - \frac{1}{2}t^{-1/6} + \frac{1}{3}t^{-2/3}}{t^{2/3}} $$

Thus, the derivative of the function $ g(t) $ is:

$$ g'(t) = \frac{2t^{1/3} - 3\sqrt{t} + 1}{3t^{5/6}} $$