"Proof of a Formula for Differentiating x^p with Respect to y

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hedlund
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Let y = x^p where p is a natural number. Is it true that
[tex]\frac{dx^n}{d^ny} = \frac{p!}{(p-n)!} \cdot x^{p-n}[/tex] with the restriction that we define [tex](-n)! \equiv \infty[/tex] for n=1,2,3... I found this formula and I believe that it is true if we define [tex](-n)![/tex] to equal [tex]\infty[/tex].
 
on Phys.org


The formula provided is indeed correct, as long as we define (-n)! to equal infinity for all natural numbers n. This definition allows us to extend the factorial function to negative numbers, which is necessary for the formula to hold true.

To prove this formula, we can use the definition of the derivative as the limit of the difference quotient. Let's start by taking the derivative of y = x^p with respect to x:

\frac{dy}{dx} = p \cdot x^{p-1}

Next, we can rewrite this as:

\frac{dy}{dx} = \frac{p}{1} \cdot x^{p-1}

From this, we can see that the coefficient of x^{p-1} is p, which is the same as the coefficient in the formula provided. Now, let's take the nth derivative of y with respect to x:

\frac{d^n y}{dx^n} = \frac{p}{1} \cdot \frac{p-1}{2} \cdot \frac{p-2}{3} \cdot ... \cdot \frac{p-(n-1)}{n} \cdot x^{p-n}

We can rewrite this as:

\frac{d^n y}{dx^n} = \frac{p!}{(p-n)!} \cdot x^{p-n}

This matches the formula provided, but with one key difference - the restriction that (-n)! equals infinity for all natural numbers n. This is necessary because as n increases, the denominator in the expression for the nth derivative becomes larger and larger, approaching infinity. Therefore, we must define (-n)! to equal infinity in order for the formula to hold true for all values of n.

In conclusion, the formula \frac{dx^n}{d^ny} = \frac{p!}{(p-n)!} \cdot x^{p-n} is correct for y = x^p, as long as we define (-n)! to equal infinity for all natural numbers n. This formula can be useful in solving problems involving derivatives of functions with a variable exponent, such as x^p.