1. - / 1.53 points consider the following. $f(t)=t\\sin (t)$ $g(t)=1 + t$ find $f^{prime}(t)$ and…

Explanation:

Step1: Find ( f^{\prime}(t) )

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = t), (u^\prime=1), (v=\sin(t)), (v^\prime=\cos(t)). (f^{\prime}(t)=(t)^\prime\sin(t)+t(\sin(t))^\prime=\sin(t)+t\cos(t))

Step2: Find ( g^{\prime}(t) )

Use the sum rule ((u + v)^\prime=u^\prime+v^\prime), where (u = 1), (u^\prime = 0), (v=t), (v^\prime=1). (g^{\prime}(t)=(1)^\prime+(t)^\prime=0 + 1=1)

Step3: Find ( y^{\prime} )

Use the quotient rule (\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u=t\sin(t)), (u^\prime=\sin(t)+t\cos(t)) (from Step1), (v = 1 + t), (v^\prime=1) (from Step2). [ \begin{align*} y^{\prime}&=\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}} \end{align*} ]

Answer:

(f^{\prime}(t)=\sin(t)+t\cos(t)) (g^{\prime}(t)=1) (y^{\prime}=\frac{\sin(t)+t\cos(t)+t^{2}\cos(t)}{(1 + t)^{2}})