How do you differentiate #g(x) =x^2cscx# using the product rule?

1 Answer
Feb 13, 2016

Answer:

#(dg(x))/dx=2xcscx-x^2cotxcscx#

Explanation:

The product rule states that the derivative of two functions of #x#, let's call them #u# and #v#, is #u'v+uv'#.

First, let's begin by letting #u=x^2# and #v=cscx#, so our example fits with the product rule. Now we can differentiate #u# and #v#:
#u=x^2->u'=2x#
#v=cscx->v'=-cotxcscx#

On to substitutions:
#(dg(x))/dx=(u)'(v)+(u)(v)'#
#(dg(x))/dx=2xcscx-x^2cotxcscx#

This result is fine, but many teachers like the result expressed in sines and cosines. Using some identities, we can do that:
#=(2x)/sinx-(x^2cosx)/sinx*1/sinx#

#=(2x)/sinx-(x^2cosx)/sin^2x#

#=(2xsinx)/sin^2x-(x^2cosx)/sin^2x#

#=(2xsinx-x^2cosx)/sin^2x#