How do you differentiate #y = x^(-sin(x) )#?

Question

1099 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-08T08:04:21

Using the fact that

#d/dx[u^v]=u^v * [(u'v+uv'ln(u))/(u)]#

We let

#u = x# and #v = -sin(x)#

So

#y' = x^(-sin(x)) * [(-sin(x) - xcos(x) ln(x))/(x)]#

The formula we've used is

#d/dx[u^v]=u^v * [(u'v+uv'ln(u))/(u)]#