How do you find the derivative of #g(z)=(z^2+3)e^z#?

Question

1780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-09T07:55:27

#g'(z)=(z^2+2z+3)e^z#

we will use the product rule

#g(z)=u(z)v(z)=>g'(z)=color(red)(u'(z))v(z)+u(z)color(blue)(v'(z))#

we have

#g(z)=(z^2+3)e^z#

#u(z)=z^2+3=>color(red)(u'(z)=2z)#

#v(z)=e^z=>v'(z)=color(blue)(e^z)#

#:.g'(z)=color(red)(2z)e^z+(z^2+3)color(blue)(e^z)#

simplifying and tidying up

#g'(z)=e^z(2z+z^2+3)#

#g'(z)=(z^2+2z+3)e^z#