Yatayyat
Badges: 14
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
How would I go about to working out this question

Name:  Screenshot 2020-01-25 at 18.34.23.png
Views: 9
Size:  44.6 KB

Made an attempt at it using implicit differentiation, where I used the product rule too. Just want to see if this is a right way of doing it and would reach a correct answer

Name:  IMG_3731.jpg
Views: 7
Size:  501.5 KB

Note that I also used a standard result of what dy/dx would be of y = ax , so that I didn't need to write the extra steps in my working out above

Name:  IMG_3730.jpg
Views: 12
Size:  502.0 KB

Thanks!
Last edited by Yatayyat; 3 weeks ago
0
reply
Yatayyat
Badges: 14
Rep:
?
#2
Report Thread starter 3 weeks ago
#2
(Original post by vbzl)
Looks very good to me.
Ok cheers 👍
0
reply
RichE
Badges: 15
Rep:
?
#3
Report 3 weeks ago
#3
(Original post by Yatayyat)
Ok cheers 👍
Your differentiation of y^x with respect to x doesn't look right. If you set z = y^x

then lnz = x lny

so 1/z dz/dx = lny + x/y dy/dx

and then go from there...
1
reply
Yatayyat
Badges: 14
Rep:
?
#4
Report Thread starter 3 weeks ago
#4
(Original post by RichE)
Your differentiation of y^x with respect to x doesn't look right. If you set z = y^x

then lnz = x lny

so 1/z dz/dx = lny + x/y dy/dx

and then go from there...
Name:  CABA1464-E038-48BB-A433-DDB84B21C8F8.jpg.jpeg
Views: 7
Size:  14.4 KB

So is the expression I need shown above when I differentiate y^x wrt x
0
reply
ghostwalker
  • Study Helper
Badges: 17
#5
Report 3 weeks ago
#5
(Original post by Yatayyat)
Name:  CABA1464-E038-48BB-A433-DDB84B21C8F8.jpg.jpeg
Views: 7
Size:  14.4 KB

So is the expression I need shown above when I differentiate y^x wrt x
Yes, that's correct.
0
reply
Yatayyat
Badges: 14
Rep:
?
#6
Report Thread starter 3 weeks ago
#6
(Original post by ghostwalker)
Yes, that's correct.
Thanks, so I followed up from that and got my new dy/dx to be this:

Name:  IMG_3733.JPG
Views: 6
Size:  114.0 KB
0
reply
ghostwalker
  • Study Helper
Badges: 17
#7
Report 3 weeks ago
#7
(Original post by Yatayyat)
Thanks, so I followed up from that and got my new dy/dx to be this:

Name:  IMG_3733.JPG
Views: 6
Size:  114.0 KB
Yes. And you can combine the y and y^x in the numerator into y^{x+1}
1
reply
simon0
Badges: 13
Rep:
?
#8
Report 3 weeks ago
#8
Also, for part a, you need to double check your differentiation.
2
reply
Yatayyat
Badges: 14
Rep:
?
#9
Report Thread starter 3 weeks ago
#9
(Original post by simon0)
Also, for part a, you need to double check your differentiation.
Have I made a mistake when doing my product rule differentiation?
I forget to multiply by y I think

My new answer for part a) gives this now

Name:  Screenshot 2020-01-26 at 18.39.19.jpg
Views: 5
Size:  31.9 KB
0
reply
RDKGames
Badges: 20
Rep:
?
#10
Report 3 weeks ago
#10
(Original post by Yatayyat)
Have I made a mistake when doing my product rule differentiation?
I forget to multiply by y I think

My new answer for part a) gives this now
Yep
0
reply
Yatayyat
Badges: 14
Rep:
?
#11
Report Thread starter 3 weeks ago
#11
(Original post by RDKGames)
Yep
Thanks
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Bournemouth University
    Undergraduate Open Day Undergraduate
    Wed, 19 Feb '20
  • Buckinghamshire New University
    Postgraduate and professional courses Postgraduate
    Wed, 19 Feb '20
  • University of Warwick
    Warwick Business School Postgraduate
    Thu, 20 Feb '20

Have you ever accessed mental health support services at University?

Yes (44)
24.18%
No, but I have thought about it (49)
26.92%
No (89)
48.9%

Watched Threads

View All