differentiate.\n$y = \\sec(\\theta)\\tan(\\theta)$\n$y=$\n\n2. - / 1 points\nfind the derivative of the…

Explanation:

Step1: Recall product - rule

The product - rule states that if (y = u\cdot v), then (y'=u'v + uv'). For (y=\sec(\theta)\tan(\theta)), let (u = \sec(\theta)) and (v=\tan(\theta)).

Step2: Find derivatives of (u) and (v)

The derivative of (\sec(\theta)) is (\sec(\theta)\tan(\theta)), and the derivative of (\tan(\theta)) is (\sec^{2}(\theta)).

Step3: Apply product - rule

(y'=\sec(\theta)\tan(\theta)\cdot\tan(\theta)+\sec(\theta)\cdot\sec^{2}(\theta)). Simplify to get (y'=\sec(\theta)\tan^{2}(\theta)+\sec^{3}(\theta)=\sec(\theta)(\tan^{2}(\theta)+\sec^{2}(\theta))).

For (f(t)=t\cos(t)+t^{2}\sin(t)):

Step1: Apply sum - rule of differentiation

If (y = g(t)+h(t)), then (y'=g'(t)+h'(t)). Here (g(t)=t\cos(t)) and (h(t)=t^{2}\sin(t)).

Step2: Apply product - rule to (g(t))

Let (u = t) and (v=\cos(t)). Then (u' = 1) and (v'=-\sin(t)). By product - rule (g'(t)=1\cdot\cos(t)+t\cdot(-\sin(t))=\cos(t)-t\sin(t)).

Step3: Apply product - rule to (h(t))

Let (u = t^{2}) and (v=\sin(t)). Then (u' = 2t) and (v'=\cos(t)). By product - rule (h'(t)=2t\sin(t)+t^{2}\cos(t)).

Step4: Find (f'(t))

(f'(t)=g'(t)+h'(t)=\cos(t)-t\sin(t)+2t\sin(t)+t^{2}\cos(t)=\cos(t)+t\sin(t)+t^{2}\cos(t)).

Answer:

For (y = \sec(\theta)\tan(\theta)), (y'=\sec(\theta)(\tan^{2}(\theta)+\sec^{2}(\theta))) For (f(t)=t\cos(t)+t^{2}\sin(t)), (f'(t)=\cos(t)+t\sin(t)+t^{2}\cos(t))