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

1 Answer

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.

;)