Question

7-52: Find the derivative of the function

12. \(F\left( t \right) = {\left( {\frac{1}{{2t + 1}}} \right)^4}\)

Short answer

The derivative of the function is \(F'\left( t \right) = - \frac{8}{{{{\left( {2t + 1} \right)}^5}}}\).

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:

\(\begin{aligned}F\left( t \right) &= {\left( {\frac{1}{{2t + 1}}} \right)^4}\\ &= {\left( {{{\left( {2t + 1} \right)}^{ - 1}}} \right)^4}\\ &= {\left( {2t + 1} \right)^{ - 4}}\end{aligned}\)

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

\(\begin{aligned}F'\left( t \right) &= \frac{d}{{dt}}{\left( {2t + 1} \right)^{ - 4}}\\ &= - 4{\left( {2t + 1} \right)^{ - 5}} \cdot \frac{d}{{dt}}\left( {2t + 1} \right)\\ &= - 4{\left( {2t + 1} \right)^{ - 5}} \cdot \left( 2 \right)\\ &= - \frac{8}{{{{\left( {2t + 1} \right)}^5}}}\end{aligned}\)

Thus, the derivative of the function is \(F'\left( t \right) = - \frac{8}{{{{\left( {2t + 1} \right)}^5}}}\).