Question

Establish the following rules for working with differentials (where \(c\) denotes a constant and \(u\)and\(v\) are functions of \(x\)).

(a) \(dc = 0\) (b) \(d\left( {cu} \right) = cdu\)

(c) \(d\left( {u + v} \right) = du + dv\) (d) \(d\left( {uv} \right) = udv + vdu\)

(e) \(d\left( {\frac{u}{v}} \right) = \frac{{vdu - udv}}{{{v^2}}}\) (f) \(d\left( {{x^n}} \right) = n{x^{n - 1}}dx\)

Short answer

1. The rule \(dc = 0\) is proved.
2. The rule \(dcu = cdu\) is proved.
3. The rule \(d\left( {u + v} \right) = dv + du\) is proved.
4. The rule \(d\left( {uv} \right) = udv + vdu\) is proved.
5. The rule \(d\left( {\frac{u}{v}} \right) = \frac{{vdu - udv}}{{{v^2}}}\) is proved.
6. The rule \(d\left( {{x^n}} \right) = n{x^{n - 1}}dx\) is proved.

Step-by-step solution

Prove the rule in (a)

Multiply and divide by\(dx\)on the both sides of \(dc\) as follows:

^.\(dc = dx\left( {\frac{{dc}}{{dx}}} \right)\)

The derivative of constant is equal to 0. So, \(\frac{{dc}}{{dx}} = 0\). On plugging \(\frac{{dc}}{{dx}} = 0\) in the above equation, we get:

\(\begin{aligned}{c}dc &= dx\left( 0 \right)\\dc &= 0\end{aligned}\)

Prove the rule in (b)

Multiply and divide by\(dx\)on the both sides of \(d\left( {cu} \right)\) as follows:

^.\(dcu = dx\left( {\frac{{dcu}}{{dx}}} \right)\)

The constant can be taken out of the derivative. So, the above equation can be simplified as:

\(\begin{aligned}{c}dcu &= dx\left( {\frac{{dcu}}{{dx}}} \right)\\ &= dx\left( {c\frac{{du}}{{dx}}} \right)\\ &= c\left( {dx\frac{{du}}{{dx}}} \right)\\ &= cdu\end{aligned}\)

Prove the rule in (c)

Multiply and divide\(dx\)on the both sides of \(d\left( {u + v} \right)\) as follows:

^.\(d\left( {u + v} \right) = \frac{d}{{dx}}\left( {u + v} \right)dx\)

On simplifying the above equation, we get:

\(\begin{aligned}{c}d\left( {u + v} \right) &= \frac{d}{{dx}}\left( {u + v} \right)dx\\ &= \left( {\frac{{dv}}{{dx}} + \frac{{du}}{{dx}}} \right)dx\\ &= \frac{{dv}}{{dx}}dx + \frac{{du}}{{dx}}dx\\ &= dv + du\end{aligned}\)

Prove the rule in (d)

Multiply and divide\(dx\)on the both sides of \(d\left( {uv} \right)\) as follows:

\(\begin{aligned}{c}d\left( {uv} \right) &= \frac{d}{{dx}}\left( {uv} \right)dx\\ &= \left( {v\frac{{du}}{{dx}} + u\frac{{dv}}{{dx}}} \right)dx\end{aligned}\)

On simplifying the above equation, we get:

\(\begin{aligned}{c}d\left( {uv} \right) &= \left( {v\frac{{du}}{{dx}} + u\frac{{dv}}{{dx}}} \right)dx\\ &= u\frac{{dv}}{{dx}}dx + v\frac{{du}}{{dx}}dx\\ &= udv + vdu\end{aligned}\)

Prove the rule in (e)

Multiply and divide\(dx\)on the both sides of \(d\left( {\frac{u}{v}} \right)\) as follows:

\(\begin{aligned}{c}d\left( {\frac{u}{v}} \right) &= \frac{d}{{dx}}\left( {\frac{u}{v}} \right)dx\\ &= \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}\end{aligned}\)

On simplifying the above equation, we get:

\(\begin{aligned}{c}d\left( {\frac{u}{v}} \right) &= \left( {\frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}} \right)dx\\ &= \frac{{v\frac{{du}}{{dx}}dx - u\frac{{dv}}{{dx}}dx}}{{{v^2}}}\\ &= \frac{{vdu - udv}}{{{v^2}}}\end{aligned}\)

Prove the rule in (f)

Multiply and divide\(dx\)on the both sides of \(d\left( {{x^n}} \right)\)and simplify as follows:

\(\begin{aligned}{c}d\left( {{x^n}} \right) &= \frac{d}{{dx}}\left( {{x^n}} \right)dx\\ &= n{x^{n - 1}}dx\end{aligned}\):