Derivative of y=logex (1 Viewer)

[untitled....]

Hey guys, I just cant seem to get the hang of these qus.

Find the maxima, minima or pts of inflexion for the following functions and skectch a graph for each.

a) y= xlnx-x
b) y= lnx/x
c) y= logx+ 1/x

You dont have to do all of them, but any help would be great

 

[untitled....]

Thanks heaps!
And duuude how the hell do you teach yourself 4 unit; smart much?

 

[addikaye03]

Thanks heaps!
And duuude how the hell do you teach yourself 4 unit; smart much?

Thats alright, anytime u have any Q feel free to direct them my way and haha nah im not that smart, just committed.

 

[addikaye03]

c) y= logx+ 1/x

y=logx+x^-1

dy/dx=1/x-(x)^-2

dy/dx=1/x-1/x^2...S.P occurs when d/dx=0

1/x-1/x^2=0

(x^2-x)/x^3=0.. this can be simplified to (x-1)/x^2=0

so, when multiplying the denominator x-1=0, therefore x=1

when x=1, y=In1+1/1=1 therefore S.P at (1,1)

TEST gradient change. Since when x=1, dy/dx=0

When x=0.5, dy/dx<0

When x=1.5,dy/dx>0

Therefore the min T.P at (1,1)

Inflection: dy/dx=(x)^-1-(x)^-2

d^2y/dx^2=-(x)^-2+2(x)^-3

d^2y/dx^2=-1/x^2+2/x^3... Infl. occurs when d^2y/dx^2=0

2/x^3-1/x^2=0

(2x^2-x^3)/x^5=0

(2-x)/x^3=0...therefore possoble inflection at x=2

Check Concavity.. when x=2, d^2y/dx^2=0

when x=1.5, d^2y/dx^2>0

when x=2.5, d^2y/dx^2<0

Concavity changes theredore inflection point at (2,0)

I'm feeling generous tonight and you seem like a nice person. Enjoy

 

[roadrage75]

a) y=xInx-x,

u=x, v=Inx
u'=1, v'=1/x

dy/dx=vu'-uv'/v^2 + d/dx(-x)
=[Inx-x(1/x)]/(inx)^2 -1
=(Inx-1)/Inx^2-1

S.P occurs when dy/dx=0, (Inx-1)/Inx^2-1=0
Inx-1-Inx^2/Inx^2=0

Inx-1-Inx^2=0
In(x/x^2)-1=0
In(1/x)=1
1/x=e
therefore x=1/e aprrox ( 0.3678...)

When x=1/e, dy/dx=0, when x=0.3, dy/dx<0 and when x=0.4, dy/dx>0
So, x=1/e is a minimum T.P. Y-value ( resub): y=(1/e)In(1/e)-(1/e)=1/e[In(1/e)-1]
So the minimum T.P is (1/e, 1/e(In1/e-1))

For the inflection its the same process but u must check concavity change. Type the equation into the program "graphmatica" to see what graph it yields.

um...i really can't see what you're doing.. are you using the quotient differentiating rule, cas that wont work

if y = xlnx -x

then using product rule (uv' + vu')
y' = x(1/x) + 1*lnx - 1 = Inx

and y' = 0 when x = 1... testing close points to see whether max or minimum turning point, and we find that it is a minimum turning point.

oh and y'' = 1/x, and can never equal 0. ergo, no poitns of inflexion

 

[jet]

Yeah, you're right. They were meant to use the product rule. It is somewhat of an easy mistake to make though. My maths teacher was telling me once that, in 2u when asked to differentiate lnx, they did a product rule with ln and x.

 

[addikaye03]

um...i really can't see what you're doing.. are you using the quotient differentiating rule, cas that wont work

if y = xlnx -x

then using product rule (uv' + vu')
y' = x(1/x) + 1*lnx - 1 = Inx

and y' = 0 when x = 1... testing close points to see whether max or minimum turning point, and we find that it is a minimum turning point.

oh and y'' = 1/x, and can never equal 0. ergo, no poitns of inflexion

u are right i have absolutely no idea why i done it that way lol.. its very obviously product rule lol thats prob the stupidest mistake i have ever made.