Question

Reciprocal rule If g is differentiable, the Reciprocal Rule says that

\(\frac{{\bf{d}}}{{{\bf{d}}x}}\left( {\frac{{\bf{1}}}{{g\left( x \right)}}} \right) = - \frac{{g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^{\bf{2}}}}}\)

(a)Use the Quotient rule to prove the Reciprocal rule.

(b)Use the Reciprocal Rule to differentiate the function in Exercise 14.

(c)Use the Recirpocal Rule to verify that the Power Rule is valid for negative integers, that is,

\(\frac{{\bf{d}}}{{{\bf{d}}x}}\left( {{x^{ - n}}} \right) = - n{x^{ - n - {\bf{1}}}}\)

For all positive integers n.

Short answer

(a) The Reciprocal rule is proved using the quotient rule.

(b) The derivative of \(F\left( x \right)\) is \( - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\).

(c) The power rule is valid for all integers n.

Step-by-step solution

(a) Step 1: Prove Reciprocal Rule using Quotient Rule

The equation for the Quotient rule is \(\left( {\frac{f}{g}} \right)' = \frac{{gf' - fg'}}{{{{\left( g \right)}^2}}}\).

Differentiate the expression \(\frac{1}{{g\left( x \right)}}\) using the Quotient Rule.

\(\begin{aligned}\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{1}{{g\left( x \right)}}} \right) &= \frac{{g\left( x \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( 1 \right) - 1\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ &= \frac{{0 - g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ &= - \frac{{g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

Hence, the reciprocal rule is proved using the Quotient Rule.

(b) Step 2: Differentiate the function \(F\left( x \right) = \frac{{\bf{1}}}{{{\bf{2}}{x^{\bf{3}}} - {\bf{6}}{x^{\bf{2}}} + {\bf{5}}}}\) using reciprocal rule

Let, for the function \(F\left( x \right) = \frac{1}{{2{x^3} - 6{x^2} + 5}}\), \(g\left( x \right) = 2{x^3} - 6{x^2} + 5\).

Differentiate the function \(g\left( x \right)\).

\(\begin{aligned}g'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {2{x^3} - 6{x^2} + 5} \right)\\ &= 6{x^2} - 12x\end{aligned}\)

The derivative of \(F\left( x \right)\) using a reciprocal rule is:

\(\begin{aligned}F'\left( x \right) &= - \frac{{g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ &= - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\end{aligned}\)

So, the derivative of \(F\left( x \right)\) is \( - \frac{{6{x^2} - 12x}}{{{{\left( {2{x^3} - 6{x^2} + 5} \right)}^2}}}\).

(c) Step 3: Find the derivative of \(\frac{{\bf{d}}}{{{\bf{d}}x}}\left( {{x^{ - n}}} \right) =  - n{x^{ - n - {\bf{1}}}}\)

Let \(g\left( x \right) = {x^n}\). Then derivative of \(g\left( x \right)\) is:

\(\begin{aligned}g\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^n}} \right)\\ &= n{x^{n - 1}}\end{aligned}\)

The derivative of \({x^{ - n}}\) can be calculated as;

\(\begin{aligned}\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^{ - n}}} \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{1}{{g\left( x \right)}}} \right)\\ &= - \frac{{g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ &= - \frac{{n{x^{n - 1}}}}{{{{\left( {{x^n}} \right)}^2}}}\\ &= - \frac{{n{x^{n - 1}}}}{{{x^{2n}}}}\\ &= - n{x^{ - n - 1}}\end{aligned}\)

Hence proved, the power rule is valid for all integers n.