11. find the following\na. $\\frac{d}{dx}x^{2}\\sin(x)$\nb. $\\frac{d}{dx}e^{x}\\cos(x)$\nc…

Explanation:

Step1: Apply the product rule for differentiation

The product rule is ((uv)^\prime = u^\prime v+uv^\prime). For (y = x^{2}\sin(x)), let (u = x^{2}) and (v=\sin(x)). (u^\prime=\frac{d}{dx}(x^{2}) = 2x) and (v^\prime=\frac{d}{dx}(\sin(x))=\cos(x)). Then (\frac{d}{dx}(x^{2}\sin(x))=2x\sin(x)+x^{2}\cos(x)).

Step2: Apply the product rule for (y = e^{x}\cos(x))

Let (u = e^{x}) and (v=\cos(x)). (u^\prime=\frac{d}{dx}(e^{x}) = e^{x}) and (v^\prime=\frac{d}{dx}(\cos(x))=-\sin(x)). By the product rule ((uv)^\prime=u^\prime v + uv^\prime), we have (\frac{d}{dx}(e^{x}\cos(x))=e^{x}\cos(x)-e^{x}\sin(x)=e^{x}(\cos(x)-\sin(x))).

Step3: Apply the product rule for (y=\sec\theta\tan\theta)

Let (u = \sec\theta) and (v=\tan\theta). (u^\prime=\frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta) and (v^\prime=\frac{d}{d\theta}(\tan\theta)=\sec^{2}\theta). By the product rule ((uv)^\prime=u^\prime v+uv^\prime), we get (\frac{d}{d\theta}(\sec\theta\tan\theta)=\sec\theta\tan^{2}\theta+\sec^{3}\theta=\sec\theta(\tan^{2}\theta+\sec^{2}\theta)). Using the identity (\tan^{2}\theta + 1=\sec^{2}\theta), then (\tan^{2}\theta+\sec^{2}\theta = 2\sec^{2}\theta - 1), and (\frac{d}{d\theta}(\sec\theta\tan\theta)=\sec\theta(2\sec^{2}\theta - 1))

Step4: Apply the quotient rule

The quotient rule is (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). For (y=\frac{t\sin t}{1 + t}), let (u=t\sin t) and (v = 1 + t). First, find (u^\prime) using the product rule: (u^\prime=\sin t+t\cos t) and (v^\prime = 1). Then (\frac{d}{dt}\left(\frac{t\sin t}{1 + t}\right)=\frac{(\sin t+t\cos t)(1 + t)-t\sin t\times1}{(1 + t)^{2}}=\frac{\sin t+t\sin t+t\cos t+t^{2}\cos t - t\sin t}{(1 + t)^{2}}=\frac{\sin t+t\cos t+t^{2}\cos t}{(1 + t)^{2}})

Answer:

a. (2x\sin(x)+x^{2}\cos(x)) b. (e^{x}(\cos(x)-\sin(x))) c. (\sec\theta(2\sec^{2}\theta - 1)) d. (\frac{\sin t+t\cos t+t^{2}\cos t}{(1 + t)^{2}})