How can I prove that if y=(x^n)logx, then yn+1=n!/x?

I assume that, by ‘yn+1’, you mean the (n+1)th derivative of y with respect to x - this is often written (with the parentheses) as a superscript, e.g.

y(n+1) 

In other words, you want to prove that:

ddxn+1(xnln(x))=n!x 

I suggest that you edit the question to make your meaning clearer as, from the two answers submitted before mine, they didn’t understand you.

This doesn’t really count as a proof, but I think its a way to demonstrate why this is true.

From the product rule for differentiation,

dydx=xnddxln(x)+ln(x)ddxxn 

=xnx+nxn−1ln(x)=xn−1+nxn−1ln(x) 

Having differentiated once, we still have to differentiate a further n times.

dn+1ydxn+1=dndxn(xn−1+nxn−1ln(x)) 

From the sum rule for differentiation, we can split this into two parts:

dndxnxn−1+ndndxnxn−1ln(x) 

Let’s look at the first part. When we differentiate an expression than contains a term that is a power of x, we reduce the power by 1. So, if we differentiate xb b times, we end up with a constant [ xb−b=x0 ], and if we differentiate again, we get a zero. In this case, we’re wanting to differentiate xn−1 n times, this means that the term eventually becomes zero. So, our problem simplifies to:

ndndxnxn−1ln(x)=ndn−1dxn−1(ddxxn−1ln(x)) 

Applying the product rule again:

=ndn−1dxn−1(xn−2+(n−1)xn−2ln(x)) 

Applying the sum rule again, the first term again reduces to zero with continued differentiation, so we are left with:

ndn−1dxn−1(n−1)xn−2ln(x) 

As (n−1) is a constant, we can move it outside the differentiation; I’ll also introduce the notation n[2]=n!(n−2)! 

So, we have:

n[2]dn−1dxn−1xn−2ln(x) 

Applying the product rule again:

n[2]dn−2dxn−2(xn−3+(n−2)xn−3ln(x)) 

Applying the sum rule again, the first term again reduces to zero with continued differentiation, so we are left with:

n[3]dn−2dxn−2xn−3ln(x) 

Continuing the pattern we get:

n[4]dn−3dxn−3xn−4ln(x) 

n[5]dn−4dxn−4xn−5ln(x) 

…

n[n−1]d2dx2x1ln(x) 

n[n]ddxln(x)=n[n]x 

As the coefficient is just n! , our answer is:

n!x