Question

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

\\( r = 1 - \theta ^ { 4 } \sin \theta \\)

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

Explanation:

Step1: Differentiate the constant term

The derivative of a constant \(1\) with respect to \(\theta\) is \(0\). So, \(\frac{d}{d\theta}(1) = 0\).

Step2: Apply the product rule

The product rule states that if \(u = \theta^{4}\) and \(v=\sin\theta\), then \(\frac{d}{d\theta}(uv)=u'v + uv'\).

• Find \(u'\): \(\frac{d}{d\theta}(\theta^{4})=4\theta^{3}\) (using the power rule \(\frac{d}{d\theta}(\theta^{n})=n\theta^{n - 1}\)).
• Find \(v'\): \(\frac{d}{d\theta}(\sin\theta)=\cos\theta\).
• Then \(\frac{d}{d\theta}(\theta^{4}\sin\theta)=4\theta^{3}\sin\theta+\theta^{4}\cos\theta\).

Step3: Combine the results

Since \(r = 1-\theta^{4}\sin\theta\), \(\frac{dr}{d\theta}=\frac{d}{d\theta}(1)-\frac{d}{d\theta}(\theta^{4}\sin\theta)\).
Substitute the values from Step1 and Step2: \(\frac{dr}{d\theta}=0-(4\theta^{3}\sin\theta+\theta^{4}\cos\theta)=-\theta^{3}(4\sin\theta+\theta\cos\theta)\).

Answer:

\(-\theta^{3}(4\sin\theta+\theta\cos\theta)\)