How do you derive #y = 1/x# using the quotient rule?

Question

Aurora Alvarado · Accepted Answer

See the explanation section below.

The quotient rule tells us that the derivative of #f/g# is #(f'g-fg')/g^2# Applying this to #1/x# we note #f(x) = 1#, so #f'(x) = 0#, and #g(x) = x#, so g'(x) = 1# #d/dx(1/x) = ((0)(x) - (1) (1))/x^2 = -1/x^2#

Luke Alvarez · Answer

undefined To derive $ y = \frac{1}{x} $ using the quotient rule, where $ y $ is a function of $ x $, follow these steps:

1. Identify the numerator and denominator of the function $ y = \frac{1}{x} $. In this case, the numerator is 1 and the denominator is $ x $.
2. Apply the quotient rule, which states that for functions $ u(x) $ and $ v(x) $, the derivative of $ \frac{u(x)}{v(x)} $ is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $.
3. Let $ u(x) = 1 $ and $ v(x) = x $.
4. Find the derivatives $ u'(x) $ and $ v'(x) $. Since $ u(x) = 1 $, its derivative is $ u'(x) = 0 $. And $ v(x) = x $, its derivative is $ v'(x) = 1 $.
5. Substitute the values into the quotient rule formula: $ \frac{(0)(x) - (1)(1)}{x^2} = -\frac{1}{x^2} $.
6. Simplify the expression to get the derivative: $ -\frac{1}{x^2} $.
So, the derivative of $ y = \frac{1}{x} $ with respect to $ x $ is $ y' = -\frac{1}{x^2} $.