Help wid differn when to take logs of both sides

[ankur29]

Homework Statement

solve via newt raphson iterative method to 4dp

Find the solution x= 1/(1+x^x) to 4dp beginning at x=1

Homework Equations

Now the formula for this is

x1= x0 – f(x0)/f’(x0)

Now what I did was rearrange so f(x)= x( 1+x^x)-1=0

I need to differentiatie this expression

So say I called it y= x( 1+x^x)-1

Could I proceed by taking logs of both sides


i.e (an example where I saw people take logs of boths ides was when y=x^x and therefore lny=x^x) is that only for when there is one term equal to another

The Attempt at a Solution

how I attempted it was as follows

y= x + x^(x+1) -1

lny = lnx +(x+1)lnx - ln1

lny= lnx +xlnx +lnx –ln1

1/y dy/dx= 1/x + lnx +1

dy/dx = (x + x^(x+1) -1) (1/x + lnx +1)

now this feels wrong to me 

 

[reckon]

y=x(1+x^x)-1=f(x)=0 => ln(y)=ln(0) ?

why not differentiate f(x)=x(1+x^x)-1 and use the log method to differentiate just x^x part?

 

[ankur29]

y=x(1+x^x)-1=f(x)=0 => ln(y)=ln(0) ?

why not differentiate f(x)=x(1+x^x)-1 and use the log method to differentiate just x^x part?

the f(x) simpliefies to x + x^(x+1) -1

do you mean differentiate the x^(x+1) seperately

i don't think it can work like that?

do i not need something to equal something before i can apply teh log thing

 

[reckon]

yes, to differentiate the x^(x+1)
f(x)=x+x^(x+1)-1

let y=x^(x+1)
apply the log trick to y to get y' and you will get f'(x) as
f'(x)=1+y'