Differentiating logarithms help (1 Viewer)

[findsome1]

Hey, im stuck on this question.. can anyone help ?

Find the stationary point on the curve y=ln x/x and determine its nature.

I get stuck after using the quotient rule and getting (1-lnx)/x^2 = 0 (for stationary point), can anyone tell me how to do it properly.

Answer (e, 1/e) maximum

Thanks

 

[SpiralFlex]

Substitute into your original equation,

 

[Drongoski]

should not use "Ln" for "ln".

 

[SpiralFlex]

should not use "Ln" for "ln".

Okay.

 

[Amogh]

Substitute into your original equation,

Following on from that, remember to determine the stationary point's nature as well
This can be done by observing the sign of the first derivate on both sides of the stationary point.

So in this particular case:

We first determine the sign of the first derivative before e
Lets take, say, 1
At x = 1, the first derivative is positive

At x = e, the first derivative is zero

And now, after e
At x = 3, the first derivative is negative

Hence, the stationary point is a maximum point as the gradient changes from positive to negative.

 

[findsome1]

Yea thanks I get everything, but I dont understand mathematically how

x=e when ln (x) =1

 

[SpiralFlex]

Yea thanks I get everything, but I dont understand mathematically how

x=e when ln (x) =1

can be rewritten as

So when we substitute in,

It equals

Which from recalling the laws of logarithms is 1.

Or you can see from putting into index form.



Clearly the answer is 1.

 

[findsome1]

Ahh yea, I undestand now thx

 

[OmmU]

Remember that ln and e 'undo' each other.

ln (x) = 1
ln(e) = 1

It is just a memory thing that ln and e undo each other.

Applying this:
When you have y' = (1 - lnx) / x^2
You should focus on making the top 0. Regardless of what is on the bottom, if the top = 0 the whole fraction will = 0.

1 - lnx

By remembering that ln and e undo each other to make 1, it is easy


EDIT: Dam spiralflex too fast yet again

 

[Drongoski]

Remember that ln and e 'undo' each other.

That's what an inverse function is all about.

e^x and ln(x) are inverse functions of each other. So when you apply an inverse function on a function you "edit undo" what the function does.

f ^-1 f(x) = x

f { f^-1(x) } = x

Thus: ln(e^3.098 ) = 3.098

and e^ln(3.098) = 3.098

 

[HSCAREA]

omfg this qs is from mif ex 4.7 qs 10 :O

do you by any chance know how to do qs 9 by any chance? Find the point of inflexion on the curve y = xloge(x) - x^2

NB: that is log BASE e

I've tried the product rule to get y' = 1 + ln(x) - 2x

For stationary points y' = 0

therefore: 1 + ln(x) - 2x = 0

2x = 1 - ln(x)

x = (1 /2) - (ln(x) / 2)

now what? I'm lost as what to do next?

 

[AAEldar]

omfg this qs is from mif ex 4.7 qs 10 :O

do you by any chance know how to do qs 9 by any chance? Find the point of inflexion on the curve y = xloge(x) - x^2

NB: that is log BASE e

I've tried the product rule to get y' = 1 + ln(x) - 2x

For stationary points y' = 0

therefore: 1 + ln(x) - 2x = 0

2x = 1 - ln(x)

x = (1 /2) - (ln(x) / 2)

now what? I'm lost as what to do next?

Points of inflexion occur when the second derivative is equal to zero. So using this, and hopefully you know how to differentiate the functions, we can get the answer.

 

[HSCAREA]

thank you