Question

Find \(y^{\prime \prime}\) if \(y=\theta \tan \theta\)

Short answer

The second derivative, \(y^{\prime \prime}\), is \(2 \sec^2(\theta) + 2 \theta \sec(\theta) \tan(\theta)\)

Step-by-step solution

Find the First Derivative

To find the first derivative, we will use the product rule, since \(y= \theta \cdot \tan(\theta)\) is essentially a product of two functions, \(u(\theta)= \theta\) and \(v(\theta)= \tan(\theta)\). The product rule is \( (uv)^{\prime} = u^{\prime}v + uv^{\prime}\). Here, \(u^{\prime} = 1\) and \(v^{\prime} = \sec^2(\theta)\). So, \(y^{\prime} = 1 \cdot \tan(\theta) + \theta \cdot \sec^2(\theta) = \tan(\theta) + \theta \sec^2(\theta) \)

Find the Second Derivative

The second derivative, \(y^{\prime \prime}\), can be found by differentiating \(y^{\prime}\) which involves differentiating a sum of two functions. This also involves using both the product rule and the chain rule. First, find the derivative of \(\tan(\theta)\), which is \(\sec^2(\theta)\). Secondly, find the derivative of \(\theta \sec^2(\theta)\) using product rule, and note that the derivative of \(\sec(\theta)\) is \(\sec(\theta)\tan(\theta)\) using chain rule. So, we have \(y^{\prime \prime} = \sec^2(\theta) + \sec^2(\theta) + \theta \cdot 2 \sec(\theta) \tan(\theta) = 2 \sec^2(\theta) + 2 \theta \sec(\theta) \tan(\theta)\).