How I do I prove the Quotient Rule for derivatives?

Question

Samantha Adkison · Accepted Answer

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#

Ezekiel Alexander · Answer

undefined To prove the Quotient Rule for derivatives, follow these steps:

1. Start with the quotient of two functions: $ f(x) = \frac{g(x)}{h(x)} $.

2. Let $ u(x) = g(x) $ and $ v(x) = \frac{1}{h(x)} $. Rewrite the quotient as $ f(x) = u(x)v(x) $.

3. Use the Product Rule to find $ f'(x) = u'(x)v(x) + u(x)v'(x) $.

4. Substitute $ u(x) = g(x) $ and $ v(x) = \frac{1}{h(x)} $ into the expression.

5. Differentiate $ u(x) $ and $ v(x) $ with respect to $ x $.

6. Simplify the expression obtained in step 3 using the derivatives of $ u(x) $ and $ v(x) $.

7. Rewrite the expression in terms of $ g(x) $ and $ h(x) $ to obtain the Quotient Rule for derivatives:

$$ \frac{d}{dx}\left(\frac{g(x)}{h(x)}\right) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} $$