How do you find the second derivative of #y=2sin3x-5sin6x#?

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Answer 1 · 2015-02-23T23:18:35

First find the first derivative using the chain rule.

#dy/dx=6cos(3x)-30cos(6x) #

Now take the derivative again to find the second derivative

#(d^2y)/dx^2=-18sin(3x)+180sin(6x) #

Answer 2 · 2015-02-23T23:22:31

Hello,

Answer #y'' = -18 sin(3x) + 180 sin(6x)#.

First, you calculate the derivative :
#y' = 2 cos(3x)\times 3 - 5 cos(6x)\times 6#.

I used #d/dx sin(nx) = cos(nx)\times n#.

You can simplify :
#y' = 6 cos(3x) - 30 cos(6x)#.
Now, you calculate the second derivative :
#y'' = -6 sin(3x) \times 3 + 30 sin(6x)\times 6#

I used #d/dx cos(nx) = -sin(nx)\times n#.

You can also simplify to obtain the result.