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Quotient Rule (f/g)' = (f'g - fg')/g2
Here, we can do this by using the definition of the derivative or along with Logarithmic Definition.
Here we do the proof using Logarithmic Differentiation. We will first call the quotient y and get the log of both sides and utilize a property of logs onto the right side.
y = f(x)/g(x)
In y = In (f(x)/g(x))
= In f(x) - In g(x)
After that, we take the derivative of both sides and resolve for y′.
y'/y = (f'(x)/f(x)) - (g'(x)/g(x)) => y' = y (f'(x)/f(x)) - (g'(x)/g(x))
After that, plug into y and do some simplification to find the quotient rule.
y' = (f(x)/g(x))(f'(x)/f(x)) - (g'(x)/g(x))
= (f'(x)/g(x)) - ((g'(x) f(x))/(g(x)2))
= ((f'(x) g(x))/g(x)2) - ((f(x) g'(x))/(g(x)2))
= (f'(x) g(x) - f(x) g'(x))/ g(x)2
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