How do you find the derivative of #sin ^2 (2x) + sin (2x+1) #?
You can find it like this:
Using the chain rule in steps:
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To find the derivative of ( \sin^2(2x) + \sin(2x+1) ), we use the chain rule and the derivative of sine function:

Derivative of ( \sin^2(2x) ): [ \frac{d}{dx}(\sin^2(2x)) = 2\sin(2x)\cos(2x) \cdot 2 = 4\sin(2x)\cos(2x) ]

Derivative of ( \sin(2x+1) ): [ \frac{d}{dx}(\sin(2x+1)) = 2\cos(2x+1) ]
Therefore, the derivative of ( \sin^2(2x) + \sin(2x+1) ) is: [ 4\sin(2x)\cos(2x) + 2\cos(2x+1) ]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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