Question

Differentiate with respect to t.. y = b cos t + t2 sin t

Answer 1

We are given with the expression y = b cos t + t2 sin t and is asked to differentiate the function in terms of t. Based in the power law and law of products, we apply these to the given expresssion. the derivative is y' = -b sin t + t^2 * (cos t) + sin t * 2t. This is equal to y' = sin t (-b +2t) + t^2 cos t.

Answer 2

`(dy)/(dt)=\sin t(-b+2t)+t^2\cos t`

Explanation:

Given : Equation
`y=b\cos t+t^2\sin t`

To find : Differentiate with respect to t ?

Solution :

Equation
`y=b\cos t+t^2\sin t`

Differentiate with respect to t,

`(dy)/(dt)=(d(b\cos t+t^2\sin t))/(dt)`

`(dy)/(dt)=(d(b\cos t))/(dt)+(d(t^2\sin t))/(dt)`

We know,
`(d(\cos x))/(dx)=-\sin x`

Applying product rule,
`(d)/(dy)(u\cdot v) =u'v+v'u`

`(dy)/(dt)=-b\sin t+(d(t^2))/(dt)* \sin t+(d(\sin t))/(dt)* t^2`

`(dy)/(dt)=-b\sin t+2t* \sin t+\cos t* t^2`

`(dy)/(dt)=-b\sin t+2t\sin t+t^2\cos t`

`(dy)/(dt)=\sin t(-b+2t)+t^2\cos t`

Therefore,
`(dy)/(dt)=\sin t(-b+2t)+t^2\cos t`