Question

7-52: Find the derivative of the function

11. \(g\left( t \right) = \frac{1}{{{{\left( {2t + 1} \right)}^2}}}\)

Short answer

The derivative of the function is \(g'\left( t \right) = - \frac{4}{{{{\left( {2t + 1} \right)}^3}}}\).

Step-by-step solution

The Power rule combined with the Chain Rule

Power rule is defined as:

\(\frac{d}{{dx}}\left( {{u^n}} \right) = n{u^{n - 1}}\frac{{du}}{{dx}}\)

Additionally, \(\frac{d}{{dx}}{\left( {g\left( x \right)} \right)^n} = n{\left( {g\left( x \right)} \right)^{n - 1}} \cdot g'\left( x \right)\).

Find the derivative of the function

Rewrite the function as the power of \(t\) as shown below:

\(\begin{aligned}g\left( t \right) &= \frac{1}{{{{\left( {2t + 1} \right)}^2}}}\\ &= {\left( {2t + 1} \right)^{ - 2}}\end{aligned}\)

Use the power rule combined with the chain rule to obtain the derivative of the function as shown below:

\(\begin{aligned}g'\left( t \right) &= \frac{d}{{dt}}{\left( {2t + 1} \right)^{ - 2}}\\ &= - 2{\left( {2t + 1} \right)^{ - 3}} \cdot \frac{d}{{dt}}\left( {2t + 1} \right)\\ &= - 2{\left( {2t + 1} \right)^{ - 3}} \cdot \left( 2 \right)\\ &= - \frac{4}{{{{\left( {2t + 1} \right)}^3}}}\end{aligned}\)

Thus, the derivative of the function is \(g'\left( t \right) = - \frac{4}{{{{\left( {2t + 1} \right)}^3}}}\).