Question

find \\( \frac { d r } { d \theta } \\).

\\( r = ( 7 + \sec \theta ) \sin \theta \\)

\\( \frac { d r } { d \theta } = \\)

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 7+\sec\theta\) and \(v=\sin\theta\). First, find \(u^\prime\) and \(v^\prime\).
The derivative of \(u = 7+\sec\theta\) with respect to \(\theta\) is \(u^\prime=\sec\theta\tan\theta\) (since \(\frac{d}{d\theta}(7) = 0\) and \(\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta\)), and the derivative of \(v=\sin\theta\) with respect to \(\theta\) is \(v^\prime=\cos\theta\).

Step2: Substitute into the product rule formula

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Simplify \(\sec\theta\tan\theta\sin\theta\):
Since \(\sec\theta=\frac{1}{\cos\theta}\) and \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), then \(\sec\theta\tan\theta\sin\theta=\frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}\cdot\sin\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}=\tan^{2}\theta\).
Also, \((7+\sec\theta)\cos\theta = 7\cos\theta + 1\) (because \(\sec\theta\cos\theta = 1\)).

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Answer:

\(\sec^{2}\theta + 7\cos\theta\)