Question

Find the derivative of \(f\left( x \right) = \left( {{\bf{1}} + {\bf{2}}{x^{\bf{2}}}} \right)\left( {x - {x^{\bf{2}}}} \right)\) in two ways: by using the Product Rule and by performing the multiplication first. Do your answer agree?

Short answer

By using product rule: \(f'\left( x \right) = 1 - 2x + 6{x^2} - 8{x^3}\)

By multiplication first: \(f'\left( x \right) = 1 - 2x + 6{x^2} - 8{x^3}\)

The expression of derivative is same with both the method.

Step-by-step solution

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

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

Apply product rule for the function \(f\left( x \right) = \left( {1 + 2{x^2}} \right)\left( {x - {x^2}} \right)\).

\(\begin{aligned}f'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\left( {1 + 2{x^2}} \right)\left( {x - {x^2}} \right)} \right)\\ &= \left( {1 + 2{x^2}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x - {x^2}} \right) + \left( {x - {x^2}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {1 + 2{x^2}} \right)\\ &= \left( {1 + 2{x^2}} \right)\left( {1 - 2x} \right) + \left( {x - {x^2}} \right)\left( {1 + 4x} \right)\\ &= 1 + 2{x^2} - 2x - 4{x^3} + x - {x^2} + 4{x^2} - 4{x^3}\\ &= 1 - 2x + 6{x^2} - 8{x^3}\end{aligned}\)

Therefore, \(f'\left( x \right) = 1 - 2x + 6{x^2} - 8{x^3}\) by product rule.

Simplify the function \(f\left( x \right) = \left( {{\bf{1}} + {\bf{2}}{x^{\bf{2}}}} \right)\left( {x - {x^{\bf{2}}}} \right)\)

Thefunction\(f\left( x \right) = \left( {1 + 2{x^2}} \right)\left( {x - {x^2}} \right)\) can be simplified as,

\(\begin{aligned}f\left( x \right) &= \left( {1 + 2{x^2}} \right)\left( {x - {x^2}} \right)\\ &= x - {x^2} + 2{x^3} - 2{x^4}\end{aligned}\)

Differentiate the function \(f\left( x \right) = x - {x^{\bf{2}}} + {\bf{2}}{x^{\bf{3}}} - {\bf{2}}{x^{\bf{4}}}\)

Differentaite the function \(f\left( x \right) = x - {x^2} + 2{x^3} - 2{x^4}\).

\(\begin{aligned}f\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x - {x^2} + 2{x^3} - 2{x^4}} \right)\\ &= 1 - 2x + 6{x^2} - 8{x^3}\end{aligned}\)

Therefore, \(f'\left( x \right) = 1 - 2x + 6{x^2} - 8{x^3}\) by multiplication first.

Thus, both the methods agrees to derivative for same answer.