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Section 3.9 : Chain Rule
24. Differentiate \(f\left( t \right) = {\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)^3}\) .
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Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative.
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
\[f'\left( t \right) = 3{\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)^2}\frac{d}{{dt}}\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)\] Show Step 2In this step we can see that we’ll need to use the Chain Rule on each of the terms.
The derivative is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( t \right) = 3{{\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)}^2}\left( { - 6{{\bf{e}}^{ - 6t}} - \cos \left( {2 - t} \right)} \right)}}\]