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Chapter 3: Problem 48

Find \(y^{\prime \prime}\) for the following functions. $$y=\cos \theta \sin \theta$$

Short Answer

Expert verified
Answer: The second derivative is \(y^{\prime \prime} = -2\sin^2 \theta - 2\cos^2 \theta\).

Step by step solution

01

Find the first derivative of the function

We are given the function \(y = \cos \theta \sin \theta\). To find the first derivative, we will use the product rule, which states that the derivative of a product of two functions is \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of \(\theta\). In this case, let \(u = \cos \theta\) and \(v = \sin \theta\). First, find the derivatives of \(u\) and \(v\) with respect to \(\theta\): $$u' = \frac{d}{d\theta}(\cos \theta) = -\sin \theta$$ $$v' = \frac{d}{d\theta}(\sin \theta) = \cos \theta$$ Now, apply the product rule to find \(y^{\prime}\): $$y^{\prime} = u'v + uv'$$ $$y^{\prime} = -\sin \theta \sin \theta + \cos \theta \cos \theta$$ $$y^{\prime} = -\sin^2 \theta + \cos^2 \theta$$
02

Find the second derivative of the function

Now we need to find the second derivative, \(y^{\prime \prime}\). To do this, we differentiate the first derivative with respect to \(\theta\). $$y^{\prime \prime} = \frac{d}{d\theta}(-\sin^2 \theta + \cos^2 \theta)$$ Using the chain rule, we have: $$y^{\prime \prime} = \frac{d}{d\theta}(-\sin^2 \theta) + \frac{d}{d\theta}(\cos^2 \theta)$$ Differentiating each term, we get: $$y^{\prime \prime} = -2\sin \theta \cos \theta + 2\cos \theta (-\sin \theta)$$ $$y^{\prime \prime} = -2\sin^2 \theta - 2\cos^2 \theta$$ The second derivative, \(y^{\prime \prime}\), of the function \(y = \cos \theta \sin \theta\) is: $$y^{\prime \prime} = -2\sin^2 \theta - 2\cos^2 \theta$$

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