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find y'' if y=cos3xsin2x,   and if x=60°
         y'=cos3x•d/dx(sin2x)+sin2x•d/dx(cos3x)
         y'=cos3x•cos2x•2+sin2x(-sin3x)•3
         y'=2(cos3x•cos2x)-3(sin3x•sin2x)

Is it correct and..
 What's next ?

Here you've been given the double derivative, y''

Your first integral is either (they're the same)

\(y'=-\dfrac{\cos\left(5x\right)-5\cos\left(x\right)}{10}+C\)

or

\(y'=\frac{\cos \left(x\right)}{2}-\frac{\cos \left(5x\right)}{10}+C\)

Where \(C\) is the constant.

Then you find the second integral, lets us the first of the two versions above to find it:

\(y=-\frac{\sin \left(5x\right)}{50}+\frac{\sin \left(x\right)}{2}+Cx+k\)

Where \(k\) is the constant.

Now substitute \(x=60\degree \) into the equation above to obtain an expression in terms of y, C, and k.

Any further instructions provided?