Question

3–34: Differentiate the function.

30. \(G\left( q \right) = {\left( {1 + {q^{ - 1}}} \right)^2}\)

Short answer

Derivative of the above function is \( - 2\left( {{q^{ - 2}} + {q^{ - 3}}} \right)\).

Step-by-step solution

Precise Definition of differentiation

The derivative is a function of the real variable. It is the rate of change of output value with respect to an input value. The derivative function is a single variable at a selected input value.

Rewrite the function

Rewrite the given function by using the identity \({\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\).

\(\begin{align}G\left( q \right) &= {\left( {1 + {q^{ - 1}}} \right)^2}\\ &= 1 + 2{q^{ - 1}} + {\left( {{q^{ - 1}}} \right)^2}\\ &= 1 + 2{q^{ - 1}} + {q^{ - 2}}\end{align}\)

Power rule of differentiation

The power rule of differentiation is that\(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), where\(n\)is any real number.

Apply power rule and simplify.

\(\begin{align}\frac{{d\left( {G\left( q \right)} \right)}}{{dq}} &= \frac{d}{{dq}}\left( {1 + 2{q^{ - 1}} + {q^{ - 2}}} \right)\\G'\left( q \right) &= \frac{d}{{dq}}\left( 1 \right) + \frac{d}{{dq}}\left( {2{q^{ - 1}}} \right) + \frac{d}{{dq}}\left( {{q^{ - 2}}} \right)\\ &= 0 + 2\left( { - 1{q^{ - 1 - 1}}} \right) + \left( { - 2{q^{ - 2 - 1}}} \right)\\ &= - 2{q^{ - 2}} - 2{q^{ - 3}}\\ &= - 2\left( {{q^{ - 2}} + {q^{ - 3}}} \right)\end{align}\)

So, the differentiation of the function is \(G'\left( q \right) = - 2\left( {{q^{ - 2}} + {q^{ - 3}}} \right)\).