find \\( \\frac { d r } { d \\theta } \\).\n\n\\( r = ( 7 + \\sec \\theta ) \\sin \\theta \\)\n\n\\( \\frac…

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

[ \begin{align*} \frac{dr}{d\theta}&=(7 + \sec\theta)^\prime\sin\theta+(7+\sec\theta)(\sin\theta)^\prime\ &=\sec\theta\tan\theta\sin\theta+(7+\sec\theta)\cos\theta \end{align*} ] 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)). [ \begin{align*} \frac{dr}{d\theta}&=\tan^{2}\theta+7\cos\theta + 1\ &=\sec^{2}\theta- 1+7\cos\theta + 1\ &=\sec^{2}\theta+7\cos\theta \end{align*} ]

Answer:

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