Question

Differentiate

\(f\left( x \right) = \left( {{\bf{3}}{x^{\bf{2}}} - {\bf{5}}x} \right){e^x}\)

Short answer

The derivative of f is \({e^x}\left( {3{x^2} + x - 5} \right)\).

Step-by-step solution

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

The equation for product rule is \(\left( {fg} \right)' = fg' + gf'\).

Apply product rule for the function \(f\left( x \right) = \left( {3{x^2} - 5x} \right){e^x}\).

\(\begin{aligned}f'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\left( {3{x^2} - 5x} \right){e^x}} \right)\\ &= \left( {3{x^2} - 5x} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {3{x^2} - 5x} \right)\end{aligned}\)

Differentiate the equation in step 1

The derivative of f can be obtained as,

\(\begin{aligned}f'\left( x \right) &= \left( {3{x^2} - 5x} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {3{x^2} - 5x} \right)\\ &= \left( {3{x^2} - 5x} \right){e^x} + {e^x}\left( {6x - 5} \right)\\ &= {e^x}\left( {3{x^2} - 5x + 6x - 5} \right)\\ &= {e^x}\left( {3{x^2} + x - 5} \right)\end{aligned}\)

Thus, the derivative of f is \({e^x}\left( {3{x^2} + x - 5} \right)\).