Answer:
Yes, you are right.
See explanation.
Step-by-step explanation:
The definition of derivative is:
.
We are given
.
Assume
are constants.
If
then
.
Let's plug them into our definition above:


I'm going to find a common denominator for the main fraction's numerator.
That is, I'm going to multiply first fraction by
and
I'm going to multiply second fraction by
.
This gives me:

Now we can combine the fractions in the numerator:

I'm going to multiply a bit on top and see if there is anything than can be canceled:

Note: I do see that
.
I also see
.
I will also distributive in other places in the mini-fraction's numerator.

Note: I see
.
I also see
.

In the numerator of the mini-fraction on top the two terms contain a factor of
so I can factor that out.
This will give me something to cancel out across the main fraction since
.


So we now have gotten rid of what would make this over 0 if we had replace
with 0.
So now to evaluate the limit, that is also we have to do now.



I'm going to go ahead and rewrite this so that isn't over 1 anymore because we don't need the division over 1.


or what you wrote:
