Question

Find the derivative of the function:

\(f\left( z \right) = {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\)

Short answer

The derivative of

\(f\left( z \right) = {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\)

Step-by-step solution

The quotient rule of differentiation

If a function \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) and \(f\), \(g\) are both differentiable, then the derivative of \(h\left( x \right)\) is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\)

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the given function

Differentiate \(f\left( z \right)\) with respect to \(z\) as follows:

\(\begin{aligned} f'\left( z \right) &= \frac{d}{{dz}}\left[ {{e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}} \right) \\ &= {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\frac{d}{{dz}}\left( {\frac{z}{{z - 1}}} \right) \\ &= {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\left( {\frac{{\left( {z - 1} \right)\frac{d}{{dz}}\left( z \right) - z\frac{d}{{dz}}\left( {z - 1} \right)}}{{{{\left( {z - 1} \right)}^2}}}} \right) \\ &= {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\left( {\frac{{\left( {z - 1} \right)\left( 1 \right) - z\left( 1 \right)}}{{{{\left( {z - 1} \right)}^2}}}} \right)\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( z \right) &= {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}\left( {\frac{{z - 1 - z}}{{{{\left( {z - 1} \right)}^2}}}} \right)\\ &= {e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right. } {\left( {z - 1} \right)}}}}\left( {\frac{{ - 1}}{{{{\left( {z - 1} \right)}^2}}}} \right)\\ &= - \frac{{{e^{{z \mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right. } {\left( {z - 1} \right)}}}}}}{{{{\left( {z - 1} \right)}^2}}}\end{aligned}\)

Hence, the derivative of \(f\left( z \right)\) is \( - \frac{{{e^{{z\mathord{\left/{\vphantom {z {\left( {z - 1} \right)}}} \right.} {\left( {z - 1} \right)}}}}}}{{{{\left( {z - 1} \right)}^2}}}\).