How do you differentiate #y=xsin(1/x)#?

1 Answer
thelittlestleafeon
Mar 5, 2018

#y'=sin(1/x)-1/xcos(1/x)#
This is a product rule + chain rule.

Explanation:

Because the inside of the sine function is something other than x, we have to do a chain rule.
So that I know what I'm doing and why, I'm going to do the chain rule first and then show how it fits into the product rule.
The derivative of #sin(1/x)# is #sin(1/x)=sin(x^-1)=cos(x^-1)(-x^-2)=cos(1/x)(-1/x^2)#

Next we'll do the product rule and put the derivative of #sin(1/x)# in there.
#y'=(1)sin(1/x)+(x)cos(1/x)(-1/x^2)#.

Now all that's left is to simplify to the final answer:
#y'=sin(1/x)-1/xcos(1/x)#.

Related questions
See all questions in Chain Rule
Impact of this question
28067 views around the world
You can reuse this answer
Creative Commons License