What is the derivative of $\frac{4}{\sqrt{x}}$ $4/sqrtx$ ?

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Answer 1 · 2015-06-02T00:51:49

Following an exponential rule that states $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ $a^(m/n)=root(n)(a^m)$ , we can rewrite the fucntion as

$\frac{4}{x^{\frac{1}{2}}}$ $4/x^(1/2)$

Another exponential rule states that $\frac{1}{a^{n}} = a^{- n}$ $1/a^n=a^-n$

Thus,

$\frac{4}{x^{\frac{1}{2}}} = 4 x^{- (\frac{1}{2})}$ $4/x^(1/2)=4x^-(1/2)$

$\frac{d y}{d x} = (- \frac{1}{2}) 4 x^{- \frac{3}{2}} = - 2 x^{- \frac{3}{2}}$ $(dy)/(dx)=(-1/2)4x^(-3/2)=-2x^(-3/2)$

Or, if you prefer the original form, we can go back to the fraction and root:

$\frac{d y}{d x} = - \frac{2}{\sqrt{x^{3}}}$ $(dy)/(dx)=-2/sqrt(x^3)$

What is the derivative of 4√x4/sqrtx?