A problem with a power rule proof

[madah12]

Homework Statement

y=xⁿ
<=>
ln(y)=nln(x)
<=>d/dx ln(y) = d/dx n ln(x) <=> y'/y=n/x
<=> y'=y*n/x=x^n*x/x=nx^(n-1)

Homework Equations

The Attempt at a Solution

but doesn't this proof only hold for positive x? because ln a negative number is undefined or am I missing something?

 

[HallsofIvy]

Yes, it does. What you could do is note that for even n, [itex](-x)^n= x^n[/itex] and for odd n, [itex](-x)^n= -x^n[/itex] to extend from positive to negative numbers. Of course, you would still have to do x= 0 separately but sinced [itex]0^n= 0[/itex], that is easy. That only proves it for the power a positive integer but x any real number.

But, personally, I would consider that proof "overkill". Lograrithms are much more advanced than polynomials so I would expect to know the derivative of [itex]x^n[/itex] long before the derivative of logarithm.

Prove [itex](x^n)'= n x^{n-1}[/itex] for positive integer n by induction on n:
If n= 1, [itex]x^n= x[/itex] and its derivative is [itex]1= 1(x^0)[/itex].

Now, suppose [itex](x^k)'= k x^{k-1}[/itex] for some k. Then [itex]x^{k+1}= (x^k)(x)[/itex] so using the product rule,
[tex](x^{k+1})'= (x^k)'(x)+ (x^k)(x')= (kx^{k-1})x+ (x^k)(1)= kx^k+ x^k= (k+1)x^k[/tex].

You could then use the quotient rule to extend [itex](x^n)'= n x^{n-1}[/itex] to negative n.

But [itex]x^r[/itex] where r is an arbitrary rational or irrational number is only defined for x positive anyway so you can use the "logarithm" proof for those.