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

1 Answer

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

We know f(x)=y=(xe)xcscx

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

dydx=ddx((xe)xcscx)

=(xe)xd(cscx)dx+cscxd((xe)x)dx

(use Product Rule):

d(uv)dx=udvdx+vdudx, where u and v are the participating variables)

d(cscx)dx=cotxcscx

So:
dydx=(xe)x(cotxcscx)+cscxd(xxex)dx

d(xxex)dx=xxd(ex)dx+exd(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(cotxcscx)+cscx(xxex+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(cotxlnx2)

which is equivalent to the second alternate form here.

;)