To find derivative of: f(x)=(xe)^x*cscxf(x)=(xe)x⋅cscx
We know f(x)=y=(xe)^x*cscxf(x)=y=(xe)x⋅cscx
Now, differentiating both side with respect to x, we get:
dy/dx=d/dx((xe)^x*cscx)dydx=ddx((xe)x⋅cscx)
= (xe)^x*(d(cscx))/dx+cscx*(d((xe)^x ))/dx=(xe)x⋅d(cscx)dx+cscx⋅d((xe)x)dx
(use Product Rule):
(d(uv))/dx=u*(dv)/dx+v*(du)/dxd(uv)dx=u⋅dvdx+v⋅dudx, where u and v are the participating variables)
(d(cscx))/dx=-cotx*cscxd(cscx)dx=−cotx⋅cscx
So:
dy/dx=(xe)^x*(-cotx*cscx)+cscx*(d(x^x*e^x ))/dxdydx=(xe)x⋅(−cotx⋅cscx)+cscx⋅d(xx⋅ex)dx
(d(x^x*e^x))/dx=x^x*(d(e^x) )/dx+e^x*(d(x^x) )/dxd(xx⋅ex)dx=xx⋅d(ex)dx+ex⋅d(xx)dx using product rule again.
Find (d(x^x))/dxd(xx)dx using logarithmic differentiation:
w = x^xw=xx so lnw = x lnxlnw=xlnx and, differentiating implicitly:
1/w (dw)/dx = lnx+11wdwdx=lnx+1 so (dw)/dx = wlnx + wdwdx=wlnx+w
and (d(x^x))/dx = x^xlnx + x^xd(xx)dx=xxlnx+xx
then:
dy/dx=(xe)^x*(-cotx*cscx)+cscx*(x^x*e^x+e^x*(x^xlnx + x^x))dydx=(xe)x⋅(−cotx⋅cscx)+cscx⋅(xx⋅ex+ex⋅(xxlnx+xx))
If we want to simplify this more, multiply things out:
= -x^xe^xcotxcscx+cscx(x^xe^x+x^xe^xlnx + x^x e^x)=−xxexcotxcscx+cscx(xxex+xxexlnx+xxex)
And keep going:
= -x^xe^xcotxcscx+cscx x^xe^x+cscx x^xe^xlnx + cscx x^x e^x=−xxexcotxcscx+cscxxxex+cscxxxexlnx+cscxxxex
Factor out e^x x^xexxx:
= x^xe^x(-cotxcscx+cscx+cscx lnx + cscx)=xxex(−cotxcscx+cscx+cscxlnx+cscx)
And then -cscx−cscx:
= -e^x x^xcscx(cotx - lnx - 2)=−exxxcscx(cotx−lnx−2)
which is equivalent to the second alternate form here.
;)