Question
Tutorial Exercise Find the derivative of the function. \[ f(t)=6 t \sin (\pi t) \] Step 1 We note that the given function \( f(t)=6 t \sin (\pi t) \) is the product of two differentiable functions of the form \( f(t)=g(t) h(t) \) where \( g(t)=6 t \) and \( h(t)=\sin (\pi t) \). However, before we can apply the product rule we must first find \( h^{\prime}(t) \). Doing so requires the use of the chain rule because \( h(t)=\sin (\pi t) \) is a composite function with \( u=\pi t \) and \( h(u)=\sin (u) \). Furthermore, we recall that, in general, if \( y=\sin (u) \), where \( u \) is a differentiable function of \( t \), then, by the chain rule, we have the following. \[ \begin{array}{l}\frac{d y}{d t}=\frac{d y}{d u} \frac{d u}{d t}=\cos (u) \frac{d u}{d t} \\ \quad u=\pi t \\ \frac{d u}{d t}=\square \\ \text { So, we first find } \frac{d u}{d t} \\ \text { SUBMIT }\end{array} \]
Ask by Sullivan Page. in the United States
Mar 15,2025
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Tutor-Verified Answer
Answer
\[
\frac{d u}{d t} = \pi
\]
Solution
\[
\frac{d u}{d t} = \pi
\]
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Mind Expander
To find \( \frac{d u}{d t} \), we differentiate \( u = \pi t \) with respect to \( t \). The derivative is simply the constant multiplied by the derivative of \( t \), which gives us: \[ \frac{d u}{d t} = \pi. \] Now, using the chain rule we previously discussed, we can find \( h^{\prime}(t) \): \[ h^{\prime}(t) = \cos(u) \cdot \frac{d u}{d t} = \cos(\pi t) \cdot \pi. \]