How do you find the derivative of the function $g (t) = \frac{1}{t^{\frac{1}{2}}}$ $g(t) = 1/t^(1/2)$ ?

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Answer 1 · 2015-12-31T18:52:00

First, we can rearrange it.

From a power rule, we know that $a^{- n} = \frac{1}{a^{n}}$ $a^-n=1/a^n$ .

Another power rule states that $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ $a^(m/n)=root(n)(a^m)$

Thus, we can rewrite $g (t) = t^{- \frac{1}{2}}$ $g(t)=t^(-1/2)$

Now, we simply differentiate it:

$\frac{d g (t)}{d t} = (- \frac{1}{2}) \cdot t^{- \frac{3}{2}}$ $(dg(t))/(dt)=(-1/2)*t^(-3/2)$

$\frac{d g (t)}{d t} = - \frac{1}{2 t^{\frac{3}{2}}}$ $(dg(t))/(dt)=-1/(2t^(3/2))$

How do you find the derivative of the function g(t) = 1/t^(1/2)g(t)=1t12g(t) = 1/t^(1/2)?