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:
#dy/dx=d/dx((xe)^x*cscx)#
#= (xe)^x*(d(cscx))/dx+cscx*(d((xe)^x ))/dx#
(use Product Rule):
#(d(uv))/dx=u*(dv)/dx+v*(du)/dx#, where u and v are the participating variables)
#(d(cscx))/dx=-cotx*cscx#
So:
#dy/dx=(xe)^x*(-cotx*cscx)+cscx*(d(x^x*e^x ))/dx#
#(d(x^x*e^x))/dx=x^x*(d(e^x) )/dx+e^x*(d(x^x) )/dx# using product rule again.
Find #(d(x^x))/dx# using logarithmic differentiation:
#w = x^x# so #lnw = x lnx# and, differentiating implicitly:
#1/w (dw)/dx = lnx+1# so #(dw)/dx = wlnx + w#
and #(d(x^x))/dx = x^xlnx + x^x#
then:
#dy/dx=(xe)^x*(-cotx*cscx)+cscx*(x^x*e^x+e^x*(x^xlnx + x^x))#
If we want to simplify this more, multiply things out:
#= -x^xe^xcotxcscx+cscx(x^xe^x+x^xe^xlnx + x^x e^x)#
And keep going:
#= -x^xe^xcotxcscx+cscx x^xe^x+cscx x^xe^xlnx + cscx x^x e^x#
Factor out #e^x x^x#:
#= x^xe^x(-cotxcscx+cscx+cscx lnx + cscx)#
And then #-cscx#:
#= -e^x x^xcscx(cotx - lnx - 2)#
which is equivalent to the second alternate form here.
;)
