Use the fundamental definition of a derivative to find f'(x) where f(x)=

the answer i get is , but i'm not entirely sure if this is correct and also i'm not sure if i'm using the right method.

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:

(b-a)/(x+b)²

Step-by-step explanation:

f(x+h) = (x+h+a)/(x+h+b)

limit h -->0

[f(x+h) - f(x)]/h

(x+h+a)/(x+h+b) - (x+a)/(x+b)

[(x+b)(x+h+a) - (x+h+b)(x+a)] ÷ [h(x+h+b)(x+b)]

[x²+xb+hx+hb+ax+ab-x²-ax-hx-ha-bx-ab] ÷ [h(x+h+b)(x+b)]

[bh - ah] ÷ [h(x+h+b)(x+b)]

h(b-a) ÷ [h(x+h+b)(x+b)]

(b-a) ÷ [(x+h+b)(x+b)]

As h --> 0

(b-a)/(x+b)²

d

step-by-step explanation: