Log Questions (1 Viewer)

[Mc_Meaney]

SO CONFUSED!!! I have no Idea in hell how to do these

(i) Derive y= log(e) (2-x)^1/2

(ii) Find the tanjent to the curve y=Log(e)x t (2, log(e)2)

(iii) Equation of the Tanjent to y= log(e)(x-1) at the point where x=2

Help Me plz plz

 

[pLuvia]

(i) Derive y= log(e) (2-x)^1/2

d/dx (ln(2-x)^1/2))
=-1/2(2-x)^-1/2/(2-x)^1/2
=-1/(4-2x)

(ii) Find the tanjent to the curve y=Log(e)x t (2, log(e)2)

Looking at this question, it seems this tangent can't even happen as ln(0) doesn't exist?

 

[Mc_Meaney]

y=Log(e)x at (2, log(e)2)

thats better...does that help?

 

[pLuvia]

Is that the question? If so then yes

y=lnx
dy/dx=1/x

At (2, ln2)
dy/dx=1/2

Tangent:
y-ln=1/2(x-2)
y=1/2x-1+ln2

(iii) Equation of the Tanjent to y= log(e)(x-1) at the point where x=2

y= ln(x-1)
dy/dx=1/(x-1)

At (2,0) dy/dx=1
tangent:
y=(x-2)

Hope that helped

 

[Mc_Meaney]

Is that the question? If so then yes

y=lnx
dy/dx=1/x

At (2, ln2)
dy/dx=1/2

Tangent:
y-ln=1/2(x-2)
y=1/2x-1+ln2

(iii) Equation of the Tanjent to y= log(e)(x-1) at the point where x=2

y= ln(x-1)
dy/dx=1/(x-1)

At (2,0) dy/dx=1
tangent:
y=(x-2)

Hope that helped

Your above average intelligence is very much appreciated.

 

[pLuvia]

Just remember though

d/dx[lnf(x)]=f'(x)/f(x)

and that's how you differentiate log equations

 

[Mc_Meaney]

Find the stat points on y=x^3.e^x and determine the nature.

Ok, so what I've got already

u= x^3 v=e^x
du/dx = 3x^2 dv/dx= 1/x

So then.... (e^x.3x^2) - (x^3.1/x)/ (1/x)^2

Am I good so far? How to I finish....so lost

 

[Riviet]

d/dx(e^x) =/= 1/x

You can either use the 2nd derivative and sub. in the x values of your stat pts or test either side of your stat pts using 1st derivative test.

 

[Mc_Meaney]

d/dx(e^x) =/= 1/x

You can either use the 2nd derivative and sub. in the x values of your stat pts or test either side of your stat pts using 1st derivative test.

Finding stat points is my prob :'(

I equate to zero yes?

 

[Riviet]

Yes, and factorising helps too. You should get:

Spoiler

x=0, -3

 

[Mc_Meaney]

OK so I've got to here:

0=3x^2e^x-x^2

0= x^2.(3e^x -1)

0= 3^x - 1

1= 3^x....am I good so far?

I can see that will = 0 but how do you get -3?

 

[Riviet]

dy/dx=x³e^x+3e^xx²=0 [using product rule]
x²e^x(x+3)=0
x=0, -3

 

[Mc_Meaney]

Thanks heaps Riv, your a champ

 

[Riviet]

It's a pleasure to help.