How do you differentiate f(x) = (xe)^x csc xf(x)=(xe)xcscx?

1 Answer

To find derivative of: f(x)=(xe)^x*cscxf(x)=(xe)xcscx

We know f(x)=y=(xe)^x*cscxf(x)=y=(xe)xcscx

Now, differentiating both side with respect to x, we get:

dy/dx=d/dx((xe)^x*cscx)dydx=ddx((xe)xcscx)

= (xe)^x*(d(cscx))/dx+cscx*(d((xe)^x ))/dx=(xe)xd(cscx)dx+cscxd((xe)x)dx

(use Product Rule):

(d(uv))/dx=u*(dv)/dx+v*(du)/dxd(uv)dx=udvdx+vdudx, where u and v are the participating variables)

(d(cscx))/dx=-cotx*cscxd(cscx)dx=cotxcscx

So:
dy/dx=(xe)^x*(-cotx*cscx)+cscx*(d(x^x*e^x ))/dxdydx=(xe)x(cotxcscx)+cscxd(xxex)dx

(d(x^x*e^x))/dx=x^x*(d(e^x) )/dx+e^x*(d(x^x) )/dxd(xxex)dx=xxd(ex)dx+exd(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(cotxcscx)+cscx(xxex+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 -cscxcscx:
= -e^x x^xcscx(cotx - lnx - 2)=exxxcscx(cotxlnx2)

which is equivalent to the second alternate form here.

;)