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Differentiate Twice

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 


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
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Associated Topics:
High School Calculus

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