Question

11-18: Find the differential of the function.

13. \(y = \frac{{1 + 2u}}{{1 + 3u}}\)

Short answer

The differential of the function is \(dy = - \frac{1}{{{{\left( {1 + 3u} \right)}^2}}}du\).

Step-by-step solution

Definition of differentials

The equation establishes the differential \(dy\)with respect to \(dx\) as shown below:

\(dy = f'\left( x \right)dx\)

Where \(dx\) represent the independent variableand \(dy\) represent the dependent variable.

Determine the differential of the function

The function is \(y = f\left( u \right) = \frac{{1 + 2u}}{{1 + 3u}}\).

Evaluate the derivative of the function as shown below:

\(\begin{aligned}f'\left( u \right) &= \frac{d}{{du}}\left( {\frac{{1 + 2u}}{{1 + 3u}}} \right)\\ &= \frac{{\left( {1 + 3u} \right)\left( 2 \right) - \left( {1 + 2u} \right)\left( 3 \right)}}{{{{\left( {1 + 3u} \right)}^2}}}\\ &= \frac{{2 + 6u - 3 - 6u}}{{{{\left( {1 + 3u} \right)}^2}}}\\ &= - \frac{1}{{{{\left( {1 + 3u} \right)}^2}}}\end{aligned}\)

Obtain the differential of the function is shown below:

\(\begin{aligned}dy &= f'\left( u \right)du\\dy &= - \frac{1}{{{{\left( {1 + 3u} \right)}^2}}}du\end{aligned}\)

Thus, the differential of the function is \(dy = - \frac{1}{{{{\left( {1 + 3u} \right)}^2}}}du\).