Question

Use the Chain Rule and the Product Rule to give an alternative proof of the Quotient Rule.

(Hint: Write \(\frac{{f\left( x \right)}}{{g\left( x \right)}} = f\left( x \right){\left( {g\left( x \right)} \right)^{ - 1}}\).)

Short answer

The answer is \(\frac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).

Step-by-step solution

Chain Rule of Derivative

Let \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) be a composition of function. The derivative of this function with respect to \(x\) is:

\(\begin{aligned}F'\left( x \right) &= \frac{d}{{dx}}f\left( {g\left( x \right)} \right)\\ &= f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\end{aligned}\)

So \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Alternate proof of Quotient Rule

Let \(F\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\). Rewriting the expression we get \(F\left( x \right) = f\left( x \right){\left( {g\left( x \right)} \right)^{ - 1}}\).

Differentiate the function with respect to \(x\) is:

\(\begin{array}{l}F'\left( x \right) = \frac{d}{{dx}}f\left( x \right){\left( {g\left( x \right)} \right)^{ - 1}}\\ = f\left( x \right)\frac{d}{{dx}}{\left( {g\left( x \right)} \right)^{ - 1}} + {\left( {g\left( x \right)} \right)^{ - 1}}f'\left( x \right)\\ = f\left( x \right)\left( { - 1} \right){\left( {g\left( x \right)} \right)^{ - 2}}g'\left( x \right) + {\left( {g\left( x \right)} \right)^{ - 1}}f'\left( x \right)\\ = \frac{{f'\left( x \right)}}{{g\left( x \right)}} - \frac{{f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ = \frac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{array}\)

Hence, \(F'\left( x \right) = \frac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).