Comment on Activity 1

For part (b), `ftext[(]x text[)]=gtext[(]x text[)]htext[(]x text[)]`, where `gtext[(]x text[)]=x` and `htext[(]x text[)]=x^2`. Since `ftext[(]x text[)]=x^3`, we know `f'text[(]x text[)]=3x^2. On the other hand, `g'text[(]x text[)]=1` and `h'text[(]x text[)]=2x`. So, it is clear that `f'text[(]x text[)]` is not `g'text[(]x text[)]h'text[(]x text[)], i.e., the obvious candidate for a product formula is wrong. Parts (a) and (c) show the same thing.

We will see shortly that the correct formula is

`f'text[(]x text[)]=gtext[(]x text[)]h'text[(]x text[)]+g'text[(]x text[)]htext[(]x text[)]`.

For example, in part (b),

`gtext[(]x text[)]h'text[(]x text[)]+g'text[(]x text[)]htext[(]x text[)]=1*x^2+x*2x=3x^2`

Let's try this out for part (c). Here, `ftext[(]x text[)]=gtext[(]x text[)]htext[(]x text[)]`, where `gtext[(]x text[)]=e^(2x)` and `htext[(]x text[)]=e^(3x)`. For this example,

`gtext[(]x text[)]h'text[(]x text[)]+g'text[(]x text[)]htext[(]x text[)]=e^(2x)*3e^(3x)+2e^(2x)*e^(3x)=5e^(5x)`.

On the other hand, `ftext[(]x text[)]=e^(5x)` and `f'text[(]x text[)]=5e^(5x)`.