# Homework Help: Integration involving ln

1. Nov 17, 2009

### xllx

1. The problem statement, all variables and given/known data
Find dy/dx of y=1/xlnx

2. Relevant equations
I thought that I'd have to use the quotient rule:
y'=vu'-uv'/v^2

However the differential of 1 is 0, so is this the right rule to use?

3. The attempt at a solution

u=1
u'=0

v=xlnx
v'=1/x

y'=vu'-uv'/v^2
y'=(xlnx*0)-(1*1/x)/(xlnx)^2
y'=-1/x(xlnx)^2
y'= -1/x^3lnx^2

But this doesnt seem right to me, I'm not sure. Any help at all would be greatly appreciated. Many Thanks.

2. Nov 17, 2009

### Dopefish1337

Really? I thought 1/x was the derivative of lnx, not xlnx.

3. Nov 17, 2009

### xllx

Yes sorry just realised that..
I redid it and got dy/dx=1-lnx/(x^2lnx^2), which sounds a more accurate answer and think I have done it right this time.

Is is possible to just extend this question though:
If I was to find when the gradient equals 0, I've got 0=1-lnx so 1=lnx, but how would I solve this?

Thankyou

4. Nov 17, 2009

### Dopefish1337

e to the what equals 1?

5. Nov 17, 2009

### xllx

Does lne=1??
But I dont know how that solves this, does that mean x=e?

6. Nov 17, 2009

### Dopefish1337

Yes, ln(e)=1.

I kindve messed up my hint there and ended up inadvertantly giving answer instead. (e^0 is 1, but thats not relevent here.)

Remember ln is just log base e, which 'undoes' powers, so you you're looking for the power of e that equals e in this case. (e^1=e^ln(x) => e^(lnx) is just x, =>e=x).