Question

Use the fundamental definition of a derivative to find f'(x) where f(x)=$\frac{x+a}{x+b}$

The answer I get is $\frac{-a+b}{\left(x+b\right)^{2}}$ , but I'm not entirely sure if this is correct and also I'm not sure if I'm using the right method.

Answer 1

Answer:

Yes, you are right.

See explanation.

Step-by-step explanation:

The definition of derivative is:

$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.

We are given $f(x)=\frac{x+a}{x+b}$.

Assume $a \text{ and } b$ are constants.

If $f(x)=\frac{x+a}{x+b}$ then $f(x+h)=\frac{(x+h)+a}{(x+h)+b}$.

Let's plug them into our definition above:

$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{(x+h)+a}{(x+h)+b}-\frac{x+a}{x+b}}{h}$

I'm going to find a common denominator for the main fraction's numerator.

That is, I'm going to multiply first fraction by $1=\frac{x+b}{x+b}$ and

I'm going to multiply second fraction by $1=\frac{(x+h)+b}{(x+h)+b}$.

This gives me:

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{((x+h)+a)(x+b)}{((x+h)+b)(x+b)}-\frac{(x+a)((x+h)+b)}{(x+b)((x+h)+b)}}{h}$

Now we can combine the fractions in the numerator:

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{((x+h)+a)(x+b)-(x+a)((x+h)+b)}{((x+h)+b)(x+b)}}{h}$

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

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{(x+h)x+(x+h)b+ax+ab-x(x+h)-xb-a(x+h)-ab}{((x+h)+b)(x+b)}}{h}$

Note: I do see that $(x+h)x-x(x+h)=0$.

I also see $ab-ab=0$.

I will also distributive in other places in the mini-fraction's numerator.

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{xb+bh+ax-xb-ax-ah}{((x+h)+b)(x+b)}}{h}$

Note: I see $xb-xb=0$.

I also see $ax-ax=0$.

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{bh-ah}{((x+h)+b)(x+b)}}{h}$

In the numerator of the mini-fraction on top the two terms contain a factor of $h$ so I can factor that out.

This will give me something to cancel out across the main fraction since $\frac{h}{h}=1$.

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{h(b-a)}{((x+h)+b)(x+b)}}{h}$

$f'(x)=\lim_{h \rightarrow 0} \frac{\frac{(b-a)}{((x+h)+b)(x+b)}}{1}$

So we now have gotten rid of what would make this over 0 if we had replace $h$ with 0.

So now to evaluate the limit, that is also we have to do now.

$\frac{\frac{(b-a)}{((x+0)+b)(x+b)}}{1}$

$\frac{\frac{b-a)}{((x)+b)(x+b)}}{1}$

$\frac{\frac{(b-a)}{(x+b)(x+b)}}{1}$

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.

$\frac{b-a}{(x+b)(x+b)}$

$\frac{b-a}{(x+b)^2}$

or what you wrote:

$\frac{-a+b}{(x+b)^2}$

Answer 2

Answer:

(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