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I've been posed a question in which I'm to differentiate with respect to $x$ a function of the form $(x+a)^k$. I've successfully completed (matches the book's answer) the question by using the chain rule, however I cannot achieve the same result using the definition of the derivative.

I would like a worked example of differentiation from first principles of a function of the form $(x+a)^k$.

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Differentiation with respect to $x$? –  user17794 Jul 7 '12 at 0:29
    
Yes. Question updated to reflect your feedback. –  kobrien Jul 7 '12 at 0:30
    
Is there any particular reason why you want it done from first principles? Things like the chain rule were proved precisely so you don't have to do everything from first principles. –  Potato Jul 7 '12 at 0:32
    
I'm aware the chain rule exists for this purpose, but it bothers me that I can't get it by first principles when it's a rather simple function. I'd like to see where I'm going wrong. It's for academic rather than practical benefit. –  kobrien Jul 7 '12 at 0:35

3 Answers 3

up vote 4 down vote accepted

For concreteness, let's first take $k=3$.

We use the factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$.

Here we have $(x+a+h)^3-(x+a)^3=h((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2)$.

The difference quotient is then

$$\frac{(x+a+h)^3-(x+a)^3}{h}=((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2).$$

As $h$ goes to zero, this becomes $3(x+a)^2$, as desired.

How does this generalize? We have similar factorizations for all $k$:

$$x^k-y^k=(x-y)(x^k+x^{k-1}y+\cdots xy^{k-1}+ y^k)$$

where there are $k$ terms in the second parenthesis. This factorization is easy to check, because the terms all telescope and cancel.

So what I did for $k=3$ can be repeated for general $k$, and when $h$ goes to zero you get $k$ terms of $(x+a)^{k-1}$ added together, which is exactly the same derivative you get using the chain rule.

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Excellent. Using this gave me the correct result. Thanks. –  kobrien Jul 7 '12 at 1:00

We want to show that for any positive integer $n$, $$\lim_{h\to 0}\frac{(x+a+h)^n -(x+a)^n}{h}=n(x+a)^{n-1}.$$ To save typing, write $w$ instead of $x+a$. So we want to show that $$\lim_{h\to0}\frac{(w+h)^n-w^n}{h}=nw^{n-1}.$$ We prove the result by induction on $n$. The base step $n=1$ is easy. We prove the induction step, that if the result holds for $n=k$, it holds for $n=k+1$.

Note that $(w+h)^{k+1}=(w+h)^k (w+h)=w(w+h)^k +h(w+h)^k$. It follows that $$(w+h)^{k+1}-w^{k+1}=w((w+h)^k -w^k) +h(w+h)^k.$$ Divide by $h$. We get $$\frac{(w+h)^{k+1}-w^{k+1}}{h}=w\frac{(w+h)^k -w^k}{h} +(w+h)^k.\tag{$1$}$$ By the induction hypothesis, $$\lim_{h\to 0}w\frac{(w+h)^k -w^k}{h}=w(kw^{k-1})=kw^k.$$ And it is clear that $$\lim_{h\to 0}(w+h)^k=w^k.$$ So the right-hand side of $(1)$ has limit $(k+1)w^k$ has $h\to 0$. This completes the induction step.

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$$ \frac{d}{dx} (x+a)^k = \lim_{h\to 0} \frac{1}{h}\left((x+h+a)^k - (x+a)^k\right) ~~. $$ The fundamental idea is, use the binomial theorem. $$ (x+h+a)^k = \sum_{j=0}^k {k \choose j} (x+a)^{k-j} h^j ~~, $$ so we notice that in the difference in the above limit, subtracting $(x+a)^k$ removes the $0^\text{th}$ term of the summation. Hence, $$ \lim_{h\to 0} \frac{1}{h}\left((x+h+a)^k - (x+a)^k\right) = \lim_{h\to 0} \frac{1}{h}\sum_{j=1}^k {k \choose j} (x+a)^{k-j} h^j $$ $$ = \lim_{h\to 0}\sum_{j=1}^k {k \choose j} (x+a)^{k-j} h^{j-1} = \lim_{h\to 0} \sum_{j=0}^{k-1} {k \choose j+1} (x+a)^{k-j-1} h^j ~~. $$ The summation expression can be thought of as a polynomial in $h$, so the only term which survives taking the limit will be when $j=0$. Hence, $$ \lim_{h\to 0} \sum_{j=0}^{k-1} {k \choose j+1} (x+a)^{k-j-1} h^j = {k \choose 1} (x+a)^{k-1} = k(x+a)^{k-1} ~~. $$ This is the simplest clean way I know of to solve this by the definition of derivative.

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