How do you find the derivative of #s=tsint#?

Answer 1

#s'(t)=sint+tcost#

This will require the product rule for derivatives.

Recall that the product rule states that given a function that is the product of two other functions,

#s(t)=f(t)*g(t)#

its derivative is

#s'(t)=f'(t)*g(t)+f(t)*g'(t)#

For this expression,

#f(t)=t# and #g(t)=sint#

So,

#s'(t)=(1)*sint+t*cost#
#s'(t)=sint+tcost#
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Answer 2

To find the derivative of ( s = t \sin(t) ), you can use the product rule:

[ \frac{d}{dt}(t \sin(t)) = t \frac{d}{dt}(\sin(t)) + \sin(t) \frac{d}{dt}(t) ]

Differentiating ( \sin(t) ) with respect to ( t ) yields ( \cos(t) ).

Differentiating ( t ) with respect to ( t ) yields ( 1 ).

So, ( \frac{d}{dt}(t \sin(t)) = t \cos(t) + \sin(t) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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