How do I simplify the limit as h approaches 0 of ((x+h)^n-x^n)/h to prove that the derivative of x^n=nx^(n-1)?
Thanks!
Another way. Let y = x + h; then y - x = h. We have $\displaystyle y^n - x^n=(y-x)(y^{n-1}+y^{n-2}x+\dots+yx^{n-2}+x^{n-1})$. Now $\displaystyle y^{n-1}+\dots+x^{n-1}\to nx^{n-1}$ as $\displaystyle y\to x$.
an "informal proof"
[(x+h)^{n} - x^{n}]/h =
(x^{n} + nx^{n-1}h + (other stuff with h^{2} terms) - x^{n})/h =
(nx^{n-1}h)/h + h^{2}(...stuff...)/h =
nx^{n-1} + h(...stuff...).
no matter what the "...stuff..." is, it is clearly not infinite at any particular x, so we have:
$\displaystyle \lim_{h \to 0} \frac{(x+h)^n - x^n}{h} = nx^{n-1} + (0)(...\text{limit of stuff}...) = nx^{n-1}$.
Plato's post makes this argument rigorous, but this gives you "the general idea".
or, you can use induction:
base case: n = 1
then $\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 = x^0$.
assume that for n = k-1:
$\displaystyle \lim_{h \to 0} \frac{(x+h)^{k-1} - x^k}{h} = (k-1)x^{k-2}$.
then:
$\displaystyle \lim_{h \to 0} \frac{(x+h)^k - x^k}{h} = \lim_{h \to 0} \frac{(x+h)(x+h)^{k-1} - (x+h)x^{k-1} + (x+h)x^{k-1} - x(x^{k-1})}{h}$
$\displaystyle = \lim_{h \to 0}\left[ (x+h)\left(\frac{(x+h)^{k-1} - x^{k-1}}{h}\right)\right] + \lim_{h \to 0}\left[x^{k-1}\left(\frac{(x+h) - x}{h}\right)\right]$
$\displaystyle = \left(\lim_{h \to 0} (x+h) \right) \left(\lim_{h \to 0} \frac{(x+h)^{k-1} - x^{k-1}}{h}\right) + \left(\lim_{h \to 0} x^{k-1} \right) \left( \lim_{h \to 0} \frac{(x + h) - x}{h} \right)$
$\displaystyle = \left(\lim_{h \to 0} (x+h) \right)((k-1)x^{k-2}) + \left(\lim_{h \to 0} x^{k-1} \right)(1)$, by our induction hypothesis, and the base case,
$\displaystyle = (x)((k-1)x^{k-2}) + x^{k-1} = (k-1)x^{k-1} + x^{k-1} = (k-1+1)x^{k-1} = kx^{k-1}$, and we are done.
No, it isn't.
Another proof, for n an integer, would be inductive: if n= 0, then $\displaystyle x^0= 1$ is a constant so its derivative is $\displaystyle 0= 0x^{0-1}$. Assume that, for some k, the derivative of $\displaystyle x^k$ is $\displaystyle kx^{k-1}$. Then we can differentiate $\displaystyle x^{k+1}= x(x^k)$ by the product rule: the derivative is $\displaystyle (x)'(x^k)+ (x)(x^k)'= 1(x^k)+ x(kx^{k-1})= x^k+ x(kx^{k-1}= (k+1)x^k$.
For k not an integer, we can use "logarithmic differentiation". If $\displaystyle y= x^k$, then $\displaystyle ln(y)= ln(x^k)= kln(x)$. Now, we need the fact that the derivative of ln(x) is 1/x (which why we typically start with integer powers of x). The derivative of the left side is $\displaystyle \frac{1}{y}\frac{dy}{dx}$ and the derivative of the right side is $\displaystyle \frac{k}{x}$. Since $\displaystyle y= x^k$, we have $\displaystyle \frac{1}{x^k}\frac{dy}{dx}= \frac{k}{x}$. From that, we have $\displaystyle \frac{dy}{dx}= \frac{k}{x}x^k= kx^{k-1}$.