Question

Differentiate.

\(h\left( r \right) = \frac{{a{e^r}}}{{b + {e^r}}}\)

Short answer

The answer is \(h'\left( r \right) = \frac{{ab{e^r}}}{{{{\left( {b + {e^r}} \right)}^2}}}\).

Step-by-step solution

Derivative of multiplication of two function

If a function is division of two function the we can differentiate the function by the following rule:

Let the function be \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) then the derivativewill be

\(\begin{aligned}h'\left( x \right) &= \frac{d}{{dx}}h\left( x \right)\\ &= \frac{d}{{dx}}\frac{{f\left( x \right)}}{{g\left( x \right)}}\\ &= \frac{{g\left( x \right)\frac{d}{{dx}} - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

Finding the derivative of the given function

The function is given by \(h\left( r \right) = \frac{{a{e^r}}}{{b + {e^r}}}\).

Differentiate the function with respect to \(r\) we get,

\(\begin{aligned}h'\left( r \right) &= \frac{d}{{dr}}\left( {\frac{{a{e^r}}}{{b + {e^r}}}} \right)\\ &= \frac{{\left( {b + {e^r}} \right)\frac{d}{{dr}}\left( {a{e^r}} \right) - a{e^r}\frac{d}{{dr}}\left( {b + {e^r}} \right)}}{{{{\left( {b + {e^r}} \right)}^2}}}\\ &= \frac{{\left( {b + {e^r}} \right)\left( {a{e^r}} \right) - a{e^r} \cdot {e^r}}}{{{{\left( {b + {e^r}} \right)}^2}}}\\ &= \frac{{a{e^r}\left( {b + {e^r} - {e^r}} \right)}}{{{{\left( {b + {e^r}} \right)}^2}}}\\ &= \frac{{ab{e^r}}}{{{{\left( {b + {e^r}} \right)}^2}}}\end{aligned}\)

Hence \(h'\left( r \right) = \frac{{ab{e^r}}}{{{{\left( {b + {e^r}} \right)}^2}}}\).