Date: 11/14/97 at 06:13:07 From: James Subject: y = xsin(3x) prove y''+9y = 6cos(3x) How do I do this? If y = xsin(3x) prove that y''+9y = 6cos(3x), where y'' is its second derivative. James
Date: 11/25/97 at 11:43:07 From: Doctor Allan Subject: Re: y = xsin(3x) prove y''+9y = 6cos(3x) Hello James, In order to find the second derivative you must differentiate twice, and in this case you use two important rules all the time: (1) Product rule: (f*g)'(x) = f'(x)*g(x) + f(x)*g'(x) (2) Chain rule: (f o g)'(x) = f'(g(x))*g'(x) I will find the first derivative, and then you can try to find the second one and see if the equation is okay. Let's say that f(x) = x, g(x) = sin(x) and h(x) = 3x Then our original equation can be written as y = x*sin(3x) = f(x)*(g o h)(x) Using (1) we get: y' = f'(x)*(g o h)(x) + f(x)*(g o h)'(x) which yields y' = 1*sin(3x) + x*sin'(3x) = sin(3x) + x*sin'(3x) In order to calculate sin'(3x) = (g o h)'(x) we use (2): sin'(3x) = g'(h(x))*h'(x) = cos(3x)*3 So the final expression for y' will be as follows: y' = sin(3x)+x*cos(3x)*3 = sin(3x)+(3x)*cos(3x) Did you understand this? If you did, try to calculate y'' by differentiating y'. If you didn't understand please let me know, and I will try a different approach. -Doctor Allan, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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