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how do you differentiate f(t)=46+2sin(2t/3)?
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i think you need to use chain rule. the 2 in front of sine is throwing me off a bit i admit. the derivative of 46 is 0. and then with chain rule. \[2\sin (2t/3)^1 \] do the outside function first, in this case sine. so using power rule, you bring the 1 infront of the 2. and then subtract 1 from the exponent. then the derivative of sin(x) is cos(x). \[1*2\cos (2t/3)\] then you multiply by the derivative of the inside (2t/3) \[(2t/3)=2t*1/3\] power makes the derivative equal 2 and so \[2\cos (2t/3)*2=4\cos (2t/3)\]
thank you!
4/3 is in front
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