How do you differentiate #y=2^xe^x# using the product rule?

2 Answers
Dec 24, 2016

Answer:

#dy/dx=e^x(2^(x-1)+2^x)#

Explanation:

Product Rule:

If #y=uv#

Then #dy/dx=uv'+vu'#

Another way to write #dy/dx# is using the prime symbol - an apostrophe - so that will be used instead. It looks like this:

#dy/dx=y'#

#y=2^xe^x#
#u = 2^x, v=e^x#

#u'=x2^(x-1)#
#v'=e^x#

#dy/dx=vu'+uv'=e^x2^(x-1)+2^xe^x=e^x(2^(x-1)+2^x)#

Dec 26, 2016

Answer:

#(dy)/(dx)=(1+ln2)2^xe^x#

Explanation:

As #y=2^xe^x#

#lny=xln2+x#

Hence #1/yxx(dy)/(dx)=ln2+1#

and #(dy)/(dx)=(1+ln2)y=(1+ln2)2^xe^x#

Using Product Rule - Using the product rule let us assume #u=2^x# and #v=e^x#

Hence, while #(dv)/(dx)=e^x#, we have #lnu=xln2# and hence

#1/u(du)/(dx)=ln2# i.e. #(du)/(dx)=2^xln2#

As #y=uv#, we have #(dy)/(dx)=u(dv)/(dx)+v(du)/(dx)#

= #2^xe^x+e^x2^xln2#

= #(1+ln2)2^xe^x#