Question

Find the derivative of the function

19. \(f\left( t \right) = {e^{at}}\sin bt\)

Short answer

The derivative of the function is \(f'\left( t \right) = {e^{at}}\left( {b\cos bt + a\sin bt} \right)\).

Step-by-step solution

The Chain Rule

For two functions, the chain rule is defined as:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

Find the derivative of the function

When \(f\) and \(g\) are both differentiable, then the product ruleis denoted by;

\(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

Use the product rule and the chain rule to obtain the derivative of the function as shown below:

\(\begin{aligned}f'\left( t \right) &= \frac{d}{{dt}}\left( {{e^{at}}\sin bt} \right)\\ &= \left( {{e^{at}}} \right) \cdot \frac{d}{{dt}}\left( {\sin bt} \right) + \sin bt \cdot \frac{d}{{dt}}\left( {{e^{at}}} \right)\\ &= {e^{at}}\left( {\cos bt} \right) \cdot \frac{d}{{dt}}\left( {bt} \right) + \sin bt\left( {{e^{at}}} \right) \cdot \frac{d}{{dt}}\left( {at} \right)\\ &= {e^{at}}\left( {\cos bt} \right) \cdot b + \sin bt\left( {{e^{at}}} \right) \cdot a\\ &= {e^{at}}\left( {b\cos bt + a\sin bt} \right)\end{aligned}\)

Thus, the derivative of the function is \(f'\left( t \right) = {e^{at}}\left( {b\cos bt + a\sin bt} \right)\).