How do you differentiate $G (x) = {(\frac{1}{2} x)}^{5}$ $G(x)=(1/2x)^5$ ?

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Answer 1 · 2016-12-05T20:01:14

$\frac{d G}{d x} = \frac{5}{32} x^{4}$ $(dG)/dx=5/32 x^4$

if $y = x^{n}$ $y=x^n$ then $\frac{d y}{d x} = n x^{n - 1}$ $dy/dx=nx^(n-1)$

In our problem
$G (x) = {(\frac{1}{2})}^{5} x^{5}$ $G(x)=(1/2)^5 x^5$
$G (x) = (\frac{1}{32}) x^{5}$ $G(x)=(1/32) x^5$

$:. (dG)/dx=(1/32)(5x^(5-1))$
$:. (dG)/dx=5/32 x^4$

How do you differentiate G(x)=(1/2x)^5G(x)=(12x)5G(x)=(1/2x)^5?