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can somebody explain how you differentiate this type of function?
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\[f(x)=\frac{ 1 }{ x-3 }\]
Quotient Rule
Granted there are multiple ways, but generally you can just use the quotient rule.
\[f(x)=(x-3)^{-1}\]
\[\frac{ d }{ dx }x^n=nx ^{n-1}\]\[f(x)=\left( x-3 \right)^{-1}\] \[f \prime \left( x \right)=-1(x-3)^{-2}=\frac{ -1 }{ \left( x-3 \right)^2 }\]
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What @sshayer said :D
Or just give them the answer lol
\[or~\frac{ d }{ dx }(\frac{ u }{ v })=\frac{ vu \prime-uv \prime }{ v^2 }\] where u and v are functions of x
\[\frac{ d }{ dx }\left( c \right)=0\] where c is a constant.
also what @royalranger said XD
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generally we use power rule.
okay I get it now, I just wasn't sure how to change the fraction
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