Question

Differentiate

\(y = \frac{{{e^x}}}{{{\bf{1}} - {e^x}}}\)

Short answer

The derivative of y is \(\frac{{{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}}\).

Step-by-step solution

Step 1:Find the derivative of f using the Quotient rule

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

Apply Quotient rule for the function \(y = \frac{{{e^x}}}{{1 - {e^x}}}\).

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

Differentiate the equation in step 1

The derivative of y can be obtained as,

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

Thus, the derivative of yis \(\frac{{{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}}\).