V.D
Badges: 0
Rep:
?
#1
Report Thread starter 6 years ago
#1
How do you answer the following questions (it says 'use standard results to differentiate

1a) 2√x
b) 3 divided by x²
c) 1 divided by 3x3
d)one thirdx3(x - 2)
e)2 divided by x3 + √x
f) 3√x + 1 divided 2x
g)2x + 3 divided by x
h)3x² - 6 divided by x
i) 2x3 + 3x divided by √x
j) x(x² - x + 2)
k) 3x²(x² + 2x)
l) (3x - 2)(4x + 1 over x)

0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#2
Report 6 years ago
#2
(Original post by V.D)
How do you answer the following questions (it says 'use standard results to differentiate

1a) 2√x
b) 3 divided by x²
c) 1 divided by 3x3
d)one thirdx3(x - 2)
e)2 divided by x3 + √x
f) 3√x + 1 divided 2x
g)2x + 3 divided by x
h)3x² - 6 divided by x
i) 2x3 + 3x divided by √x
j) x(x² - x + 2)
k) 3x²(x² + 2x)
l) (3x - 2)(4x + 1 over x)

We're not just going to give you the answers

What standard methods have you covered in class?

If you've been through the standard rules for differentiation, then all of these should be straighforward!
0
reply
Malgorithm
Badges: 6
Rep:
?
#3
Report 6 years ago
#3
(Original post by V.D)
How do you answer the following questions (it says 'use standard results to differentiate

1a) 2√x
b) 3 divided by x²
c) 1 divided by 3x3
d)one thirdx3(x - 2)
e)2 divided by x3 + √x
f) 3√x + 1 divided 2x
g)2x + 3 divided by x
h)3x² - 6 divided by x
i) 2x3 + 3x divided by √x
j) x(x² - x + 2)
k) 3x²(x² + 2x)
l) (3x - 2)(4x + 1 over x)

For the first one, you should know that \sqrt{x} can be written as x^{\frac{1}{2}}. In order to differentiate an equation in C1, you need to get all terms to be x to the power of something. Some of these you should be able to expand or rearrange to get in this form. Then just differentiate using what you should have been taught in class.
0
reply
Nuvertion
Badges: 6
Rep:
?
#4
Report 6 years ago
#4
(Original post by Malgorithm)
For the first one, you should know that \sqrt{x} can be written as x^2. In order to differentiate an equation in C1, you need to get all terms to be x to the power of something. Some of these you should be able to expand or rearrange to get in this form. Then just differentiate using what you should have been taught in class.
Did you mean x^{\frac{1}{2}}?
0
reply
victoria98
Badges: 13
Rep:
?
#5
Report 6 years ago
#5
Where applicable, use the rules of indices to make life easier, and then just do what you'd normally do to differentiate. If you're not sure how to do that you can ask, or read your textbook. We can't do your homework for you!


Posted from TSR Mobile
0
reply
V.D
Badges: 0
Rep:
?
#6
Report Thread starter 6 years ago
#6
(Original post by davros)
We're not just going to give you the answers

What standard methods have you covered in class?

If you've been through the standard rules for differentiation, then all of these should be straighforward!
well i know if y= ax to the power n (where a is a constant)
then dy/dx = anx to the power 1 minus 1

so i know x² = 2x
but i don't know what to do when there is a 3 above the x²
0
reply
V.D
Badges: 0
Rep:
?
#7
Report Thread starter 6 years ago
#7
(Original post by victoria98)
Where applicable, use the rules of indices to make life easier, and then just do what you'd normally do to differentiate. If you're not sure how to do that you can ask, or read your textbook. We can't do your homework for you!


Posted from TSR Mobile
It's not homework it's a small section of my revision
0
reply
Jordan97
Badges: 1
Rep:
?
#8
Report 6 years ago
#8
(Original post by V.D)
well i know if y= ax to the power n (where a is a constant)
then dy/dx = anx to the power 1 minus 1

so i know x² = 2x
but i don't know what to do when there is a 3 above the x²
You need to begin by getting rid of the fraction

Remember that x^-n = 1/x^n
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#9
Report 6 years ago
#9
(Original post by V.D)
well i know if y= ax to the power n (where a is a constant)
then dy/dx = anx to the power 1 minus 1

so i know x² = 2x
but i don't know what to do when there is a 3 above the x²
This is just GCSE stuff!

\dfrac{3}{x^2} = 3x^{-2}

Now you can apply your "standard rule" for differentiation.

Follow a similar process for all the others.
0
reply
Malgorithm
Badges: 6
Rep:
?
#10
Report 6 years ago
#10
(Original post by Nuvertion)
Did you mean x^{\frac{1}{2}}?
I did, yes. Saw it soon as I posted it, fixed now XD
0
reply
V.D
Badges: 0
Rep:
?
#11
Report Thread starter 6 years ago
#11
(Original post by Jordan97)
You need to begin by getting rid of the fraction

Remember that x^-n = 1/x^n
i never learnt that above rule before, can you write it in words (i don't understand ^ means)
0
reply
victoria98
Badges: 13
Rep:
?
#12
Report 6 years ago
#12
(Original post by V.D)
It's not homework it's a small section of my revision
Ahh sorry! I'm glad you're revising! People tend to start these threads for homework help. I hope you understand the work now though ..


Posted from TSR Mobile
0
reply
V.D
Badges: 0
Rep:
?
#13
Report Thread starter 6 years ago
#13
(Original post by davros)
This is just GCSE stuff!

\dfrac{3}{x^2} = 3x^{-2}

Now you can apply your "standard rule" for differentiation.

Follow a similar process for all the others.
can you show me the steps of how you came to your answer
0
reply
victoria98
Badges: 13
Rep:
?
#14
Report 6 years ago
#14
Once you get x^something. Multiply the coefficient of x by the power of x, then drop the power by one. E.g. X^2 + 2x + 6 becomes 2x + 2


Posted from TSR Mobile
0
reply
Jordan97
Badges: 1
Rep:
?
#15
Report 6 years ago
#15
(Original post by V.D)
i never learnt that above rule before, can you write it in words (i don't understand ^ means)
Does this help?
Attached files
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#16
Report 6 years ago
#16
(Original post by V.D)
can you show me the steps of how you came to your answer
You said this was revision - "revision" means "seeing again" i.e. there should be nothing unfamiliar in all this for you!

You know that \dfrac{1}{x^2} = x^{-2} from GCSE. Therefore

\dfrac{3}{x^2} = 3x^{-2}

Now apply your standard rule to differentiate this giving

(-2)(3)x^{-3} = -6x^{-3} = \dfrac{-6}{x^3}

Follow a similar process to attack all the others
0
reply
V.D
Badges: 0
Rep:
?
#17
Report Thread starter 6 years ago
#17
(Original post by Jordan97)
Does this help?
thx
0
reply
Gaiaphage
Badges: 15
Rep:
?
#18
Report 6 years ago
#18
(Original post by V.D)
can you show me the steps of how you came to your answer
That's just a fact you have to learn, you can bring a variable to the top of a fraction from the bottom (or vice versa) and flip the sign of the power
0
reply
V.D
Badges: 0
Rep:
?
#19
Report Thread starter 6 years ago
#19
(Original post by davros)
You said this was revision - "revision" means "seeing again" i.e. there should be nothing unfamiliar in all this for you!

You know that \dfrac{1}{x^2} = x^{-2} from GCSE. Therefore

\dfrac{3}{x^2} = 3x^{-2}

Now apply your standard rule to differentiate this giving

(-2)(3)x^{-3} = -6x^{-3} = \dfrac{-6}{x^3}

Follow a similar process to attack all the others
so 1 over 3x3 will equal -9x^-3
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#20
Report 6 years ago
#20
(Original post by V.D)
so 1 over 3x3 will equal -9x^-3
Are you trying to differentiate

\dfrac{1}{3x^3}

You've gone wrong in 2 places if that's the case.

It might be a good idea to talk to your Maths teacher in the morning about this exercise as you don't seem to have covered some of the basic rules properly in class
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

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (255)
56.79%
I don't have everything I need (194)
43.21%

Watched Threads

View All
Latest
My Feed