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

Answer 1

Emma Aldridge

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))#

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Answer 2

Axel Alvarez

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:

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.