Quotient rule of f(x)/g(x)
g(x)f'(x)-f(x)g'(x)/g(x)²
The quotient rule is a formula used in differential calculus to find the derivative of a function that is in the form of f(x) / g(x), that is, a quotient of two functions.
The quotient rule states that if f(x) and g(x) are two differentiable functions, then the derivative of their quotient f(x) / g(x) is given by:
\[(f(x)/g(x))’ = (f'(x)g(x) – f(x)g'(x))/(g(x))^2\]
where f'(x) and g'(x) represent the derivatives of f(x) and g(x) with respect to x, respectively.
To derive the quotient rule, we start with the definition of the derivative,
\[f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h}\]
and consider the quotient of two functions f(x) and g(x),
\[\frac{f(x)}{g(x)}\]
We can rewrite this as
\[f(x) \cdot \frac{1}{g(x)}\]
Now, we use the product rule for derivatives, which states that if f(x) and g(x) are two differentiable functions, then the derivative of their product f(x)g(x) is given by
\[(f(x)g(x))’ = f'(x)g(x) + f(x)g'(x)\]
Applying this to the expression for f(x) / g(x), we get:
\[\left(\frac{f(x)}{g(x)}\right)’ = f(x) \cdot \left(\frac{1}{g(x)}\right)’ + \left(f(x)\right)’ \cdot \frac{1}{g(x)}\]
Now, we need to find the derivative of 1/g(x). We can do this by using the chain rule, which states that if y = f(u) and u = g(x), then the derivative of y with respect to x is given by
\[\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}\]
In this case, we let f(u) = 1/u and u = g(x), so that y = 1/g(x). Then, we have
\[\left(\frac{1}{g(x)}\right)’ = \frac{d}{dx}\left(\frac{1}{g(x)}\right) = \frac{d}{dg(x)}\left(\frac{1}{u}\right) \cdot \frac{du}{dx} = -\frac{1}{(g(x))^2} \cdot g'(x)\]
Substituting this back into our equation for the derivative of f(x) / g(x), we get
\[\left(\frac{f(x)}{g(x)}\right)’ = f(x) \cdot \left(-\frac{1}{(g(x))^2} \cdot g'(x)\right) + f'(x) \cdot \frac{1}{g(x)}\]
Simplifying this expression gives us the quotient rule:
\[\left(\frac{f(x)}{g(x)}\right)’ = \frac{f'(x)g(x) – f(x)g'(x)}{(g(x))^2}\]
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