Question
We now apply the product rule. \( \begin{array}{l}(g(t) h(t))^{\prime}=g(t) h^{\prime}(t)+h(t) g^{\prime}(t) \\ \\ =6 t(\pi \cos (\pi t))+\sin (\pi t) \square \square\end{array} \) Therefore, we have the following result. \( f^{\prime}(t)=\square \)
Ask by Clarke Rodriquez. in the United States
Mar 15,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( f^{\prime}(t)=6\pi t\cos (\pi t)+6\sin (\pi t) \)
Solution
We start with the product rule for differentiation. Suppose we have
\[
f(t)=g(t)h(t)
\]
where
\[
g(t)=6t \quad \text{and} \quad h(t)=\sin (\pi t).
\]
Then the derivative is given by
\[
f'(t)=g(t)h'(t)+h(t)g'(t).
\]
First, we compute each derivative:
\[
g'(t)=6,
\]
\[
h'(t)=\pi\cos (\pi t).
\]
Now applying the product rule:
\[
f'(t)=6t\left(\pi\cos (\pi t)\right)+\sin (\pi t)\cdot 6.
\]
Simplifying the expression, we obtain:
\[
f'(t)=6\pi t\cos (\pi t)+6\sin (\pi t).
\]
Thus, the final result is
\[
f^{\prime}(t)=6\pi t\cos (\pi t)+6\sin (\pi t).
\]
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Simplify this solution Beyond the Answer
To complete the derivative using the product rule, let’s express it clearly. You have \( g(t) = 6t \) and \( h(t) = \sin(\pi t) \). For \( h'(t) \), the derivative is \( \pi \cos(\pi t) \). Therefore, applying the product rule gives us: \[ f'(t) = g(t) h'(t) + h(t) g'(t) = 6t(\pi \cos(\pi t)) + \sin(\pi t)(6) \] So the final result is: \[ f' (t) = 6 \pi t \cos(\pi t) + 6 \sin(\pi t) \] Want a quick tip? Remember to differentiate each function separately first, then multiply and sum them as the product rule states! It’s a common mistake to forget one of the derivatives or to mess up the signs with trigonometric functions. Keep an eye out for those!
