How do you differentiate #g(z) = z^2cos(z)e^(2+z)# using the product rule?

Question

1645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-10T08:59:19

#dy/dx=ze^(2+z)(zcos(z)-zsin(z)+2cos(z))#

#dy/dx=ze^(2+z)((2+z)cos(z)-zsin(z))#

Given #y=uvw#, where #u, v# and #w# are all functions of #z#, then #y'=uvw'+uv'w+u'vw#

#u=z^2#
#u'=2x#

#v=cos(z)#
#v'=-sin(z)#

#w=e^(2+z)#
#w'=d/dz[2+z]e^(2+z)=e^(2+z)#

So, #dy/dx=z^2cos(z)e^(2+z)-z^2sin(z)e^(2+z)+2zcos(z)e^(2+z)#

#dy/dx=ze^(2+z)(zcos(z)-zsin(z)+2cos(z))#

#dy/dx=ze^(2+z)((2+z)cos(z)-zsin(z))#