How I do I prove the Quotient Rule for derivatives?

1 Answer
Apr 30, 2016

We can use the product rule:

#d/dx[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)#

In order to prove the quotient rule, which states that

#d/dx[f(x)/g(x)]=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#

However, we can apply this to the product rule by writing #f(x)/g(x)# as #f(x)(g(x))^-1#.

Use the product rule on this:

#d/dx[f(x)(g(x))^-1]=f'(x)(g(x))^-1+f(x)d/dx[(g(x))^-1]#

Differentiating #d/dx[(g(x))^-1]# requires the chain rule:

#d/dx[(g(x))^-1]=-g'(x)(g(x))^-2#

Hence we obtain

#d/dx[f(x)(g(x))^-1]=f'(x)(g(x))^-1-f(x)g'(x)(g(x))^-2#

Rewriting with fractions, this becomes

#d/dx[f(x)/g(x)]=(f'(x))/g(x)-(f(x)g'(x))/(g(x))^2#

Rewriting with a common denominator, this becomes

#d/dx[f(x)/g(x)]=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#

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