Question

Differentiate

\(y = \left( {{\bf{4}}{x^{\bf{2}}} + {\bf{3}}} \right)\left( {{\bf{2}}x + {\bf{5}}} \right)\)

Short answer

The derivative of y is \(24{x^2} + 40x + 6\).

Step-by-step solution

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 \(y\left( x \right) = \left( {4{x^2} + 3} \right)\left( {2x + 5} \right)\).

\(\begin{aligned}y'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\left( {4{x^2} + 3} \right)\left( {2x + 5} \right)} \right)\\ &= \left( {4{x^2} + 3} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {2x + 5} \right) + \left( {2x + 5} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {4{x^2} + 3} \right)\end{aligned}\)

Differentiate the equation in step 1

The derivative of f can be obtained as,

\(\begin{aligned}y'\left( x \right) &= \left( {4{x^2} + 3} \right)\left( 2 \right) + \left( {2x + 5} \right)\left( {8x} \right)\\ &= 8{x^2} + 6 + 16{x^2} + 40x\\ &= 24{x^2} + 40x + 6\end{aligned}\)

Thus, the derivative of y is \(24{x^2} + 40x + 6\).