Question

find the derivative of $g(x) = \frac{1}{x^7}$.\
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$g(x) = \frac{1}{7x^8}$\
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$g(x) = \frac{7}{x^8}$\
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$g(x) = -\frac{7}{x^6}$\
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$g(x) = -\frac{7}{x^8}$

Explanation:

Step1: Rewrite original function

$g(x) = x^{-7}$

Step2: Apply power rule for derivatives

If $f(x)=x^n$, then $f'(x)=nx^{n-1}$. Here $n=-7$, so:
$g'(x) = -7x^{-7-1} = -7x^{-8}$

Step3: Rewrite to fractional form

$g'(x) = -\frac{7}{x^8}$

Answer:

$g'(x)=-\frac{7}{x^8}$