# How do you find the derivative of #x(t)=5sint+3cost#?

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To find the derivative of ( x(t) = 5\sin(t) + 3\cos(t) ), you can differentiate each term separately using the chain rule. The derivative of ( \sin(t) ) is ( \cos(t) ), and the derivative of ( \cos(t) ) is ( -\sin(t) ). Therefore, the derivative of ( 5\sin(t) ) is ( 5\cos(t) ), and the derivative of ( 3\cos(t) ) is ( -3\sin(t) ). Combining these results, the derivative of ( x(t) ) with respect to ( t ) is ( x'(t) = 5\cos(t) - 3\sin(t) ).

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To find the derivative of ( x(t) = 5 \sin(t) + 3 \cos(t) ), we will use the derivative rules for trigonometric functions.

The derivative of ( \sin(t) ) is ( \cos(t) ) and the derivative of ( \cos(t) ) is ( -\sin(t) ). Using these derivative rules, we can find the derivative of ( x(t) ) term by term.

( \frac{d}{dt}[5\sin(t)] = 5\cos(t) )

( \frac{d}{dt}[3\cos(t)] = -3\sin(t) )

Therefore, the derivative of ( x(t) ) with respect to ( t ) is:

( \frac{dx}{dt} = 5\cos(t) - 3\sin(t) )

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