Question

Differentiate.

\(h\left( w \right) = \left( {{w^2} + 3w} \right)\left( {{w^{ - 1}} - {w^{ - 4}}} \right)\)

Short answer

Derivative of the function is \(1 + 2{w^{ - 3}} + 9{w^{ - 4}}\)\(\).

Step-by-step solution

Precise Definition of differentiation

Derivative of a function used in calculus mathematics. It is the change of output value with respect to an input value. The process of finding derivatives is called differentiation.

Find the derivative of h using the Product rule

The equation for the Product rule is \({\left( {fg} \right)^\prime } = f'g + g'f\)

Apply this rule on the given function as:

\(\)\(\begin{aligned}\frac{d}{{dw}}\left( {h\left( w \right)} \right) &= \frac{d}{{dw}}\left( {\left( {{w^2} + 3w} \right)\left( {{w^{ - 1}} - {w^{ - 4}}} \right)} \right)\\ &= \left( {{w^2} + 3w} \right)\frac{d}{{dw}}\left( {{w^{ - 1}} - {w^{ - 4}}} \right) + \left( {{w^{ - 1}} - {w^{ - 4}}} \right)\frac{d}{{dw}}\left( {{w^2} + 3w} \right)\end{aligned}\)

Simplify the obtained condition

Solve this according to the formula.

\(\)\(\begin{aligned}h'\left( w \right) &= \left( {{w^2} + 3w} \right)\frac{d}{{dw}}\left( {{w^{ - 1}} - {w^{ - 4}}} \right) + \left( {{w^{ - 1}} - {w^{ - 4}}} \right)\frac{d}{{dw}}\left( {{w^2} + 3w} \right)\\ &= \left( {{w^2} + 3w} \right)\left( { - {w^{ - 2}} + 4{w^{ - 5}}} \right) + \left( {{w^{ - 1}} - {w^{ - 4}}} \right)\left( {2w + 3} \right)\\ &= \left( { - 1 - 3{w^{ - 1}} + 12{w^{ - 4}} + 4{w^{ - 3}}} \right) + \left( {2 - 2{w^{ - 3}} + 3{w^{ - 1}} - 3{w^{ - 4}}} \right)\\ &= - 1 - 3{w^{ - 1}} + 12{w^{ - 4}} + 4{w^{ - 3}} + 2 - 2{w^{ - 3}} + 3{w^{ - 1}} - 3{w^{ - 4}}\\ &= 1 + 2{w^{ - 3}} + 9{w^{ - 4}}\end{aligned}\)

So, \(h'\left( w \right) = 1 + 2{w^{ - 3}} + 9{w^{ - 4}}\)is the final answer of this derivative function.