How do you find the second derivative of #sin(2x)#?

Answer 1

Emilia Aguilar

#f''(x)=-4sin 2x#

Let #f(x)=sin 2x#.

By #(sin x)'=cos x# and Chain Rule,

#f'(x)=cos 2x cdot(2x)'=2cos 2x#

By #(cos x)'=-sin x# and Chain Rule,

#f''(x)=-2sin 2x cdot (2x)'=-4sin 2x#

I hope that this was clear.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Christopher Agresta

To find the second derivative of ( \sin(2x) ), first, find the first derivative of ( \sin(2x) ) using the chain rule:

[ \frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot \frac{d}{dx}(2x) = 2\cos(2x) ]

Now, differentiate ( 2\cos(2x) ) with respect to ( x ) using the chain rule again:

[ \frac{d}{dx}[2\cos(2x)] = -4\sin(2x) ]

So, the second derivative of ( \sin(2x) ) is ( -4\sin(2x) ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.