Video Transcript
Find the derivative of the function 𝑓 of 𝑡 is equal to 𝑒 to the power of 𝑎𝑡 times the sin of 𝑏𝑡.
The question gives us a function 𝑓 in terms of 𝑡. It wants us to find the derivative of this function. So, we want to differentiate this with respect to 𝑡. We can see that our function 𝑓 of 𝑡 is the product of two functions. It’s the product of an exponential function and a trigonometric function. And we know how to differentiate the product of two functions by using the product rule.
If we have that 𝑓 of 𝑡 is equal to the product of two functions 𝑢 of 𝑡 and 𝑣 of 𝑡, then the product rule tells us 𝑓 prime of 𝑡 is equal to 𝑣 of 𝑡 times 𝑢 prime of 𝑡 plus 𝑢 of 𝑡 times 𝑣 prime of 𝑡. So, to use the product rule, we’ll set our function 𝑢 of 𝑡 to be 𝑒 to the power of 𝑎𝑡, and we’ll set our function 𝑣 of 𝑡 to be the sin of 𝑏𝑡. To use the product rule, we’re going to need to find the expressions for 𝑢 prime of 𝑡 and 𝑣 prime of 𝑡.
Let’s start with 𝑢 prime of 𝑡. It is the derivative of 𝑒 to the power of 𝑎𝑡 with respect to 𝑡. And we know how to evaluate this derivative. We know for any constant 𝑛, the derivative of 𝑒 to the power of 𝑛𝑡 with respect to 𝑡 is 𝑛 times 𝑒 to the power of 𝑛𝑡. So, in our case, we just have 𝑛 is equal to 𝑎. So, we get that 𝑢 prime of 𝑡 is 𝑎 times 𝑒 to the power of 𝑎𝑡.
We now need to find an expression for 𝑣 prime of 𝑡. That’s the derivative of the sin of 𝑏𝑡 with respect to 𝑡. And this is a standard trigonometric derivative result. We know for any constant 𝑛, the derivative of the sin of 𝑛𝑡 with respect to 𝑡 is equal to 𝑛 times the cos of 𝑛𝑡. So, using this result with 𝑛 equal to 𝑏, we get that 𝑣 prime of 𝑡 is equal to 𝑏 times the cos of 𝑏𝑡.
We’re now ready to find an expression for our derivative of 𝑓. By the product rule, we have 𝑓 prime of 𝑡 is equal to 𝑣 of 𝑡 times 𝑢 prime of 𝑡 plus 𝑢 of 𝑡 times 𝑣 prime of 𝑡. Substituting in our expressions for 𝑢, 𝑣, 𝑢 prime, and 𝑣 prime, we get that 𝑓 prime of 𝑡 is equal to the sin of 𝑏𝑡 times 𝑎𝑒 to the power of 𝑎𝑡 plus 𝑒 to the power of 𝑎𝑡 times 𝑏 cos of 𝑏𝑡. And we can simplify this expression. We will take out our common factor of 𝑒 to the power of 𝑎𝑡. Doing this and rearranging, we get 𝑒 to the power of 𝑎𝑡 times 𝑎 sin of 𝑏𝑡 plus 𝑏 cos of 𝑏𝑡. And this is our final answer.
Therefore, we’ve shown if 𝑓 of 𝑡 is equal to 𝑒 to the power of 𝑎𝑡 times the sin of 𝑏𝑡, then 𝑓 prime of 𝑡 is equal to 𝑒 to the power of 𝑎𝑡 times 𝑎 sin of 𝑏𝑡 plus 𝑏 cos of 𝑏𝑡.
