Find f'(x) using the definition of the derivative?

Hello,

The function 𝑓 is such that for any value of 𝑥 and 𝑦,

   𝑓(𝑥 + 𝑦) = 𝑓(𝑥) + 𝑓(𝑦) + 𝑥²𝑦 + 𝑥𝑦²

And we are given that:

   Lim(ℎ→0) 𝑓(ℎ)/ℎ = 1

► Find 𝑓'(𝑥) using the definition of the derivative?

𝑓(𝑥 + 𝑦) = 𝑓(𝑥) + 𝑓(𝑦) + 𝑥²𝑦 + 𝑥𝑦²

𝑓(𝑥 + 𝑦) − 𝑓(𝑥) = 𝑓(𝑦) + 𝑥²𝑦 + 𝑥𝑦²

𝑓(𝑥 + 𝑦) − 𝑓(𝑥) = 𝑦 × [𝑓(𝑦)/𝑦 + 𝑥² + 𝑥𝑦]

[𝑓(𝑥 + 𝑦) − 𝑓(𝑥)] / 𝑦 = 𝑓(𝑦)/𝑦 + 𝑥² + 𝑥𝑦

By definition of the derivative:

𝑓'(𝑥)

   = Lim(ℎ→0) [𝑓(𝑥 + ℎ) − 𝑓(𝑥)] / ℎ

   = Lim(ℎ→0) 𝑓(ℎ)/ℎ + 𝑥² + 𝑥ℎ ←←← Because of property above

   = Lim(ℎ→0) 𝑓(ℎ)/ℎ + Lim(ℎ→0) 𝑥² + Lim(ℎ→0) 𝑥ℎ

   = 1 + 𝑥² + 0

   = 𝑥² + 1 ◄◄◄ANSWER

► Also, why do we start out by testing 𝑥 and 𝑦 = 0?

Most time when you are studying unknown functions with this kind of properties that mix addition and multiplication, certain values are interesting to compute because they can provide hints as how the function behave, and even the value of 𝑓(𝑥) at specific values of 𝑥.

Those values are generally 0 and 1 because: ◄◄◄ANSWER

   0 is the identity element of addition (𝑥+0=0+𝑥=𝑥)

   0 is the nullifying element of multiplication (𝑥×0=0×𝑥=0)

   1 is the identity element of multiplication (𝑥×1=1×𝑥=𝑥)

This is why most methods start by setting 𝑥=0, 𝑦=0, 𝑥=1, 𝑦=1 and any combination of them.

In your case, merely setting 𝑦=0

   𝑓(𝑥 + 𝑦) = 𝑓(𝑥) + 𝑓(𝑦) + 𝑥²𝑦 + 𝑥𝑦²

   𝑓(𝑥) = 𝑓(𝑥) + 𝑓(0) + 0 + 0

with the quick conclusion that 𝑓(0)=0

► how do I find a limit when there are 2 variables. The second variable y is what's confusing me!

Just don't consider the one of the variables as a variable, but a constant!

You are told that the relationship holds true for any 𝑥 or 𝑦.

So it will be true even if 𝑥 is a constant!

That's essentially what we did solving it when we write

   Lim(ℎ→0) [𝑓(𝑥 + ℎ) − 𝑓(𝑥)] / ℎ

Here we are moving variable ℎ closer and closer to 0, but 𝑥 is merely a value that do not impact on the result (at least in our case).

So when we are looking for a limit with one variable ℎ going to a value, just treat the other variable as a constant.

► BONUS ANSWER:

Since 𝑓'(𝑥)=𝑥²+1, a simple integration will yield

   𝑓(𝑥) = 𝑥³/3 + 𝑥 + 𝐶

where 𝐶 is the integration constant.

Knowing that 𝑓(0)=0 :

   𝑓(0) = 0

   0³/3 + 0 + 𝐶 = 0

   𝐶 = 0

Thus function 𝑓 is:

   𝑓(𝑥) = 𝑥³/3 + 𝑥

We can check the property is true:

𝑓(𝑥 + 𝑦)

   = (𝑥 + 𝑦)³/3 + (𝑥 + 𝑦)

   = 𝑥³/3 + 𝑥²𝑦 + 𝑥𝑦² + 𝑦³/3 + 𝑥 + 𝑦

   = 𝑥³/3 + 𝑥 + 𝑦³/3 + 𝑦 + 𝑥²𝑦 + 𝑥𝑦²

   = 𝑓(𝑥) + 𝑓(𝑦) + 𝑥²𝑦 + 𝑥𝑦²

QED

And obviously:

Lim(ℎ→0) 𝑓(ℎ)/ℎ

   = Lim(ℎ→0) (ℎ³/3 + ℎ)/ℎ

   = Lim(ℎ→0) ℎ²/3 + 1

   = Lim(ℎ→0) ℎ²/3 + Lim(ℎ→0) 1

   = 0 + 1

   = 1

QED

Regards,

Dragon.Jade :-)

PS: It reminds me of this question from 4 years ago:

https://answers.yahoo.com/question/index?qid=20151...

Rearrange given equation to (f(x+y)−f(x))/y = f(y)/y+x²+xy

Now take limit as y→0 to get f’(x) = 1+x²