Determine the derivative of 𝑓 of 𝑡 equals 𝑡 sin of five 𝜋𝑡.
We have a function in 𝑡, which is actually the product of two functions. We could say that one of the functions is 𝑡. And the other is sin of five 𝜋𝑡. So how do we find the derivative of the product of two functions? For two functions 𝑢 and 𝑣 in 𝑥, the derivative of 𝑢 times 𝑣 is 𝑢 times the derivative of 𝑣 with respect to 𝑥 plus 𝑣 times the derivative of 𝑢 with respect to 𝑥.
Now, of course, our function is in terms of 𝑡. So we change this slightly to the derivative of 𝑢𝑣 with respect to 𝑡. And this means we can let 𝑢 be equal to 𝑡 because that’s the first function in our equation. And we can let 𝑣 be equal to sin of five 𝜋𝑡 because that’s the second function of 𝑡.
We can see that we’re going to need to differentiate both of these with respect to 𝑡. If we differentiate 𝑢 with respect to 𝑡, we simply get one. But differentiating 𝑣 with respect to 𝑡 is a little bit trickier. We could use the chain rule. But we don’t need to. We can apply a general result. And that is if we differentiate sin of some constant of 𝑡, we get that constant multiplied by cos of that constant of 𝑡. So in this case, the derivative of sin of five 𝜋𝑡 is five 𝜋 multiplied by cos of five 𝜋𝑡.
All that’s left is to substitute this back into our equation for the product rule. 𝑢 multiplied by 𝑑𝑣 𝑑𝑡 is 𝑡 multiplied by five 𝜋 cos five 𝜋𝑡. And 𝑣 multiplied by 𝑑𝑢 𝑑𝑡 is sin five 𝜋𝑡 multiplied by one, which simplifies to five 𝜋𝑡 multiplied by cos five 𝜋𝑡 plus sin of five 𝜋𝑡.