How do you differentiate #sin ^2 (2x) + sin (2x+1) #?

Answer 1

Charles Anctil

The answer is #=sin(4x)+2cos(2x+1)#

We need

#sin(u(x))'=u'(x)cos(u(x))#

#sin2a=2sinacosa#

Let #y=sin^2(2x)+sin(2x+1)#

Then, differentiating with respect to #x# (applying the chain rule)

#dy/dx=2sin(2x)*cos(2x)+2cos(2x+1)#

#=sin(4x)+2cos(2x+1)#

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Answer 2

Emily Ammerman

To differentiate sin^2(2x) + sin(2x + 1), you would use the chain rule and the sum rule of differentiation. The derivative of sin^2(2x) with respect to x is 2sin(2x)cos(2x), and the derivative of sin(2x + 1) with respect to x is 2cos(2x + 1). So, the overall derivative is:

2sin(2x)cos(2x) + 2cos(2x + 1)

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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.