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How do you find the derivative of #1/2sin2x#?

Answer 1

Evelyn Agard

#(dy)/(dx)=cos2x#

The chain rule will be applied.

#(dy)/(dx)=color(red)((dy)/(du))(du)/(dx)#

#y=1/2sin2x#

#u=2x=>(du)/(dx)=2#

#color(red)(y=1/2sinu=>(dy)/(du)=1/2cosu)#

#(dy)/(dx)=color(red)(1/2cosu ) xxx2#

#(dy)/(dx)=cosu=cos2x#

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Answer 2

William Abdalla

To find the derivative of ( \frac{1}{2} \sin(2x) ), you can use the chain rule of differentiation. The derivative of ( \sin(2x) ) with respect to ( x ) is ( 2 \cos(2x) ) according to the derivative of sine function. Multiplying by the derivative of the inner function ( 2x ) with respect to ( x ), which is ( 2 ), we get ( 2 \cdot 2 \cos(2x) = 4\cos(2x) ). Therefore, the derivative of ( \frac{1}{2} \sin(2x) ) with respect to ( x ) is ( 4\cos(2x) ).

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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.