To find derivative of: f(x)=(xe)x⋅cscx
We know f(x)=y=(xe)x⋅cscx
Now, differentiating both side with respect to x, we get:
dydx=ddx((xe)x⋅cscx)
=(xe)x⋅d(cscx)dx+cscx⋅d((xe)x)dx
(use Product Rule):
d(uv)dx=u⋅dvdx+v⋅dudx, where u and v are the participating variables)
d(cscx)dx=−cotx⋅cscx
So:
dydx=(xe)x⋅(−cotx⋅cscx)+cscx⋅d(xx⋅ex)dx
d(xx⋅ex)dx=xx⋅d(ex)dx+ex⋅d(xx)dx using product rule again.
Find d(xx)dx using logarithmic differentiation:
w=xx so lnw=xlnx and, differentiating implicitly:
1wdwdx=lnx+1 so dwdx=wlnx+w
and d(xx)dx=xxlnx+xx
then:
dydx=(xe)x⋅(−cotx⋅cscx)+cscx⋅(xx⋅ex+ex⋅(xxlnx+xx))
If we want to simplify this more, multiply things out:
=−xxexcotxcscx+cscx(xxex+xxexlnx+xxex)
And keep going:
=−xxexcotxcscx+cscxxxex+cscxxxexlnx+cscxxxex
Factor out exxx:
=xxex(−cotxcscx+cscx+cscxlnx+cscx)
And then −cscx:
=−exxxcscx(cotx−lnx−2)
which is equivalent to the second alternate form here.
;)