
Employ a standard trick used in proving the chain rule:

The limit of a product is the product of limits, i.e. we can write

The rightmost limit is an exercise in differentiating

using the definition, which you probably already know is

.
For the leftmost limit, we make a substitution

. Now, if we make a slight change to

by adding a small number

, this propagates a similar small change in

that we'll call

, so that we can set

. Then as

, we see that it's also the case that

(since we fix

). So we can write the remaining limit as

which in turn is the derivative of

, another limit you probably already know how to compute. We'd end up with

, or

.
So we find that