Answer:
Yes, you are right.
See explanation.
Step-by-step explanation:
The definition of derivative is:
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/tex].
We are given [tex]f(x)=\frac{x+a}{x+b}[/tex].
Assume [tex]a \text{ and } b[/tex] are constants.
If [tex]f(x)=\frac{x+a}{x+b}[/tex] then [tex]f(x+h)=\frac{(x+h)+a}{(x+h)+b}[/tex].
Let's plug them into our definition above:
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/tex]
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{(x+h)+a}{(x+h)+b}-\frac{x+a}{x+b}}{h}[/tex]
I'm going to find a common denominator for the main fraction's numerator.
That is, I'm going to multiply first fraction by [tex]1=\frac{x+b}{x+b}[/tex] and
I'm going to multiply second fraction by [tex]1=\frac{(x+h)+b}{(x+h)+b}[/tex].
This gives me:
[tex]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}[/tex]
Now we can combine the fractions in the numerator:
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{((x+h)+a)(x+b)-(x+a)((x+h)+b)}{((x+h)+b)(x+b)}}{h}[/tex]
I'm going to multiply a bit on top and see if there is anything than can be canceled:
[tex]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}[/tex]
Note: I do see that [tex](x+h)x-x(x+h)=0[/tex].
I also see [tex]ab-ab=0[/tex].
I will also distributive in other places in the mini-fraction's numerator.
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{xb+bh+ax-xb-ax-ah}{((x+h)+b)(x+b)}}{h}[/tex]
Note: I see [tex]xb-xb=0[/tex].
I also see [tex]ax-ax=0[/tex].
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{bh-ah}{((x+h)+b)(x+b)}}{h}[/tex]
In the numerator of the mini-fraction on top the two terms contain a factor of [tex]h[/tex] so I can factor that out.
This will give me something to cancel out across the main fraction since [tex]\frac{h}{h}=1[/tex].
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{h(b-a)}{((x+h)+b)(x+b)}}{h}[/tex]
[tex]f'(x)=\lim_{h \rightarrow 0} \frac{\frac{(b-a)}{((x+h)+b)(x+b)}}{1}[/tex]
So we now have gotten rid of what would make this over 0 if we had replace [tex]h[/tex] with 0.
So now to evaluate the limit, that is also we have to do now.
[tex]\frac{\frac{(b-a)}{((x+0)+b)(x+b)}}{1}[/tex]
[tex]\frac{\frac{b-a)}{((x)+b)(x+b)}}{1}[/tex]
[tex]\frac{\frac{(b-a)}{(x+b)(x+b)}}{1}[/tex]
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.
[tex]\frac{b-a}{(x+b)(x+b)}[/tex]
[tex]\frac{b-a}{(x+b)^2}[/tex]
or what you wrote:
[tex]\frac{-a+b}{(x+b)^2}[/tex]