Question

Differentiate.

\(f\left( x \right) = \frac{{{x^2}{e^x}}}{{{x^2} + {e^x}}}\)

Short answer

The answer is \(f'\left( x \right) = \frac{{x{e^x}\left( {{x^3} + 2{e^x}} \right)}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\).

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 \(f\left( x \right) = \frac{{{x^2}{e^x}}}{{{x^2} + {e^x}}}\).

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

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {\frac{{{x^2}{e^x}}}{{{x^2} + {e^x}}}} \right)\\ &= \frac{{\left( {{x^2} + {e^x}} \right)\frac{d}{{dx}}\left( {{x^2}{e^x}} \right) - {x^2}{e^x}\frac{d}{{dx}}\left( {{x^2} + {e^x}} \right)}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\\ &= \frac{{\left( {{x^2} + {e^x}} \right)\left( {2x{e^x} + {x^2}{e^x}} \right) - {x^2}{e^x}\left( {2x + {e^x}} \right)}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\\ &= \frac{{2{x^3}{e^x} + {x^4}{e^x} + 2x{e^{2x}} + {x^2}{e^{2x}} - 2{x^3}{e^x} - {x^2}{e^x}}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\\ &= \frac{{x{e^x}\left( {{x^3} + 2{e^x}} \right)}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\end{aligned}\)

Hence \(f'\left( x \right) = \frac{{x{e^x}\left( {{x^3} + 2{e^x}} \right)}}{{{{\left( {{x^2} + {e^x}} \right)}^2}}}\).