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pav-90 [236]
3 years ago
9

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

%2Bb%7D" id="TexFormula1" title="\frac{x+a}{x+b}" alt="\frac{x+a}{x+b}" align="absmiddle" class="latex-formula">
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.

Mathematics
2 answers:
IRINA_888 [86]3 years ago
8 0

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}

xz_007 [3.2K]3 years ago
4 0

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

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Horizontal translation left 'c' units          (x,y)  changes to  (x-c  , y)

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