Question

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

Answer 1

`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)`

Answer 2

`(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`