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, (-x)^n= x^n and for odd n, (-x)^n= -x^n to extend from positive to negative numbers. Of course, you would still have to do x= 0 separately but sinced 0^n= 0, 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 x^n long before the derivative of logarithm.

Prove (x^n)&#039;= n x^{n-1} for positive integer n by induction on n:
If n= 1, x^n= x and its derivative is 1= 1(x^0).

Now, suppose (x^k)&#039;= k x^{k-1} for some k. Then x^{k+1}= (x^k)(x) so using the product rule,
(x^{k+1})&#039;= (x^k)&#039;(x)+ (x^k)(x&#039;)= (kx^{k-1})x+ (x^k)(1)= kx^k+ x^k= (k+1)x^k.

You could then use the quotient rule to extend (x^n)&#039;= n x^{n-1} to negative n.

But x^r 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.