Another Differentiation C3 question

sorry to bother all of you again but it would be great if you could help me with these questions by showing working so I can check where I've gone wrong.

1) find the equation of the tangent to the curve y=(e^(x/3))/x at the point (3, e/3)

2) find coordinates of the turning point on the curve with equation y=(e^(3x))/x, x>0, and determine the nature of this turning point. :smile:

thanks in advance

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Reply 1

insparato

y - y_1 = m(x - x_1)

, where (x1,y1) is your point on the curve where the tangent is, finding m by differentiating y

2) Find dy/dx, set it to 0 find the values of x, then find dy^2/dx^2 pop then values of x and then that tells you whether it's min/max.

Reply 2

RVNmax

insparato

y - y_1 = m(x - x_1)

nice try but not helpful to me, I actually need the solutions as I've done all this but got the wrong answer. :please:

Reply 3

insparato

RVNmax

nice try but not helpful to me, I actually need the solutions as I've done all this but got the wrong answer. :please:

Well i'm certainly not going to do this for you. It's not that difficult to differentiate exponentials.

Show me your working out, then I can see where and how you're going wrong.

Reply 4

Farhan.Hanif93

RVNmax

nice try but not helpful to me, I actually need the solutions as I've done all this but got the wrong answer. :please:

Why don't you type up your working and we point out where you went wrong? We're not being paid to spend time typing out full solutions for people...

Reply 5

RVNmax

insparato

Well i'm certainly not going to do this for you. It's not that difficult to differentiate exponentials.

Show me your working out, then I can see where and how you're going wrong.

Farhan.Hanif93

Why don't you type up your working and we point out where you went wrong? We're not being paid to spend time typing out full solutions for people...

I never said you were lol

and as you said, it's not that difficult which was why I was asking you!

here we go:

1) dy/dx = (xe^(x/3) - e^(x/3))/x^2

= (3e^(3/3) - e^(3/3))/3^2 = 2e/9

y - e/3 = (2e/9)x - (6e/9)

so y = 2ex/9 - e/3

2) dy/dx = (e^(3x)(3x-1))/x^2

(d^2y)/(dx^2) = (e^(3x)(9x^2 - 6x + 2))/x^3

these two differentials are correct so don't worry about these.

(e^(3x)(3x-1))/x^2 = 0
so x= 1/3
this is also correct

I get 3e for the y coordinate which is correct as well

sub 1/3 into (d^2y)/(dx^2) = (e^(3x)(9x^2 - 6x + 2))/x^3
gives me -81e so it would be maximum point, but it is minimum according to answers so I think my final calculation was wrong somewhere.

:wink:

Reply 6

therapist900

feel the attitude...

Ok, which other way do you know if you have a max or min?

Reply 7

Farhan.Hanif93

RVNmax

Check this line again i.e. put x=1/3 in again and see what you get.

Reply 8

ghostwalker

RVNmax

1) find the equation of the tangent to the curve y=(e^(x/3))/x at the point (3, e/3)

RVNmax

Looks like you've used the standard quotient formula, but you've got the derivatve of e^(x/3) incorrect in part in red above.

\dfrac{d}{dx}\left(e^{x/3}\right)=\dfrac{e^{x/3}}{3}

Reply 9

RVNmax

ghostwalker

Looks like you've used the standard quotient formula, but you've got the derivatve of e^(x/3) incorrect in part in red above.

\dfrac{d}{dx}\left(e^{x/3}\right)=\dfrac{e^{x/3}}{3}

thanks a lot pal :gthumb:

Reply 10

RVNmax

Farhan.Hanif93

Check this line again i.e. put x=1/3 in again and see what you get.

(d^2y)/(dx^2) = (e^(3x)(9x^2 - 6x + 2))/x^3

= (e^(3(1/3))(9(1/3)^2 - 6(1/3) + 2))/(1/3)^3
= (e(1-2+2))/(1/27)

why do we not use bodmas/bidmas/bedmas and do 1-4 :confused:

Reply 11

wookymon

Here's the solution to 1) and the way to proceed with 2). If you need any more help, show what you've done and I'll try to help.

Part Solution.gif57.5KB

Reply 12

RVNmax

therapist900

feel the attitude...

Ok, which other way do you know if you have a max or min?

I don't know any other way. :o:

Reply 13

Farhan.Hanif93

RVNmax

(d^2y)/(dx^2) = (e^(3x)(9x^2 - 6x + 2))/x^3

= (e^(3(1/3))(9(1/3)^2 - 6(1/3) + 2))/(1/3)^3
= (e(1-2+2))/(1/27)

why do we not use bodmas/bidmas/bedmas and do 1-4 :confused:

You do use bodmas but 1-2+2 = 1.
1-2+2≠-3 :p:

Reply 14

wookymon

RVNmax

I don't know any other way. :o:

I've shown you the 2 ways....

Reply 15

RVNmax

Farhan.Hanif93

You do use bodmas but 1-2+2 = 1.
1-2+2≠-3 :p:

http://www.bbc.co.uk/schools/ks3bitesize/maths/number/order_operation/revise2.shtml

lol you're right, look at the example on the bottom of the page :p:

Reply 16

Farhan.Hanif93

RVNmax

http://www.bbc.co.uk/schools/ks3bitesize/maths/number/order_operation/revise2.shtml

lol you're right, look at the example on the bottom of the page :p:

It's there to show that it was an incorrect method. Surprised you found it but then went on to make the same mistake...

Reply 17

RVNmax

wookymon

I've shown you the 2 ways....

I saw that after the post, thanks a lot for that effort :yep:

Reply 18

RVNmax

Farhan.Hanif93

It's there to show that it was an incorrect method. Surprised you found it but then went on to make the same mistake...

I search up bodmas after you I had said it originally to you, never mind. I've been doing maths the whole day anyway, I'm going off now, cheers for the help.
:woo:

Reply 19

wookymon

RVNmax

I saw that after the post, thanks a lot for that effort :yep:

It's no problem...you should find that the 2nd derivative is e/27 using the quotient and product rules. e/27 is positive, so a minimum point. Let me know if you'd like it writing out.