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2 months ago Edited: 2 months ago, Deckz
 Derivative of transcedental function 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 ? Read 105 times 1 Reply
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2 months ago
 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?
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