quotient rule fg fg - fgg2here we can do this by

Question

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.

Proof

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