# Thread: The second derivative of Y= x(Sinx)

1. ## The second derivative of Y= x(Sinx)

For this problem I already know what the answer is. I just don't know how to get to it. The question is...

Find the second derivative of Y = x(Sinx)

I know that to get it to Y' I use the product rule which would translate into
[x(sinx)' +x'(Sinx) which gets me...

Y' = x(Cosx) + (Sinx)

This is the part I am unsure of. What I did after that was
[x(Cosx)' +x'(Cosx) + (Sinx)'] Which gave me...

x(-Sinx) + (Cosx) + (Cosx)

My end result was 2(Cosx) + x(-Sinx)

The answer is suppose to be...
Y''
= 2(Cosx) - x(Sinx)

Can anyone tell me what I did wrong, and explain to me how to get the right answer?

2. Originally Posted by Brazuca
For this problem I already know what the answer is. I just don't know how to get to it. The question is...

Find the second derivative of Y = x(Sinx)

I know that to get it to Y' I use the product rule which would translate into
[x(sinx)' +x'(Sinx) which gets me...

Y' = x(Cosx) + (Sinx)

This is the part I am unsure of. What I did after that was
[x(Cosx)' +x'(Cosx) + (Sinx)'] Which gave me...

x(-Sinx) + (Cosx) + (Cosx)

My end result was 2(Cosx) + x(-Sinx)

The answer is suppose to be...
Y''
= 2(Cosx) - x(Sinx)

Can anyone tell me what I did wrong, and explain to me how to get the right answer?
$2(Cosx) + x(-Sinx) = 2cos(x) + x*-sin(x) = 2cos(x) - xsin(x)$

3. Thanks, that was the last piece of the puzzle for me to understand this and two other problems.

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# differential of xsinx

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