Find the derivative. f(t) = 3t sin? t
The Correct Answer and Explanation is:
To find the derivative of the function:
f(t) = 3t · sin²(t)
First, rewrite the function using exponent notation:
f(t) = 3t · (sin(t))²
Now apply the product rule, since the function is a product of two functions:
- First function: u = 3t
- Second function: v = (sin(t))²
The product rule is:
f'(t) = u’v + uv’
Let us differentiate each part.
- u = 3t → u’ = 3
- v = (sin(t))²
Use the chain rule:- The outer function is something squared, so its derivative is 2·(sin(t))
- The inner function is sin(t), whose derivative is cos(t)
So:
v’ = 2·sin(t)·cos(t)
Now apply the product rule:
f'(t) = 3 · (sin(t))² + 3t · 2·sin(t)·cos(t)
Simplify the expression:
f'(t) = 3(sin(t))² + 6t·sin(t)·cos(t)
Final Answer:
f'(t) = 3(sin(t))² + 6t·sin(t)·cos(t)
Explanation
To find the derivative of a function like f(t) = 3t · sin²(t), you must recognize that it is the product of two functions: one involving t and one involving sin(t). This calls for using the product rule, which is a fundamental rule in calculus for differentiating the product of two functions.
The product rule says that if f(t) = u(t) · v(t), then its derivative is f'(t) = u'(t) · v(t) + u(t) · v'(t). In this problem, u(t) is the linear function 3t, and v(t) is the trigonometric function squared, (sin(t))².
Differentiating u(t) = 3t is straightforward, giving u'(t) = 3.
For the v(t) = (sin(t))², you use the chain rule. The outer function is a square, and the inner function is sin(t). The derivative of (sin(t))² becomes 2·sin(t)·cos(t).
Putting it all together using the product rule:
f'(t) = 3·(sin(t))² + 3t·2·sin(t)·cos(t)
This simplifies to:
f'(t) = 3(sin(t))² + 6t·sin(t)·cos(t)
This expression gives the rate of change of the original function, combining both linear and trigonometric components.
