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# Differentiation watch

Given that
y=3x^2+6x^1/3+((2x^3-7)/(3rootx))
Find dy/dx.
Thank you.
2. (Original post by HiggsBoson)
Given that
y=3x^2+6x^1/3+((2x^3-7)/(3rootx))
Find dy/dx.
Thank you.
Simplify the expression for y then use the fact that derivative of

ax^n

is

nax^(n-1).
3. (Original post by RichE)
Simplify the expression for y then use the fact that derivative of

ax^n

is

nax^(n-1).
I have tried that and got the first two terms correct, but not the last two. Can you tell me what the simplified expression is so I can compare it to see where i'm going wrong please.
4. (Original post by HiggsBoson)
I have tried that and got the first two terms correct, but not the last two. Can you tell me what the simplified expression is so I can compare it to see where i'm going wrong please.
All you need to simplify it is the power rule

x^a/x^b = x^(a-b)
5. (Original post by RichE)
All you need to simplify it is the power rule

x^a/x^b = x^(a-b)
I got,
dy/dx= 6x^2 + 2/x^2/3 + (5x^3/2)/3 + 7/6x^3/2
I this correct?
6. (Original post by HiggsBoson)
I got,
dy/dx= 6x^2 + 2/x^2/3 + (5x^3/2)/3 + 7/6x^3/2
I this correct?
Last term is wrong - or I guess is right if you mean 7/(6x^(3/2))
7. (Original post by RichE)
Last term is wrong - or I guess is right if you mean 7/(6x^(3/2))
Whats wrong with it, i cant seem to find the problem?
8. (Original post by HiggsBoson)
Whats wrong with it, i cant seem to find the problem?

(7/6) * x^(3/2)

rather than

(7/6) * x^(-3/2)

which is what it should represent.

You're not bracketting terms properly.
9. Should it not be 6x instead of 6x^2 aswell?
10. 7/6x^3/2 is (7/6) * x^(-3/2) simplified, and yeah it should be 6x lol
11. (Original post by HiggsBoson)
Given that
y=3x^2+6x^1/3+((2x^3-7)/(3rootx))
Find dy/dx.
Thank you.
For the terms in the brackets you have to use the quotient rule. The 3x^2+6x^1/3 part you just differentiate separately as normal so you have 6x + 2x^-2/3 + (insert derivate you get after you do the quotient rule)

Remember quitoent rule is (u’v+uv’)/v^2
12. (Original post by The Night King)
For the terms in the brackets you have to use the quotient rule. The 3x^2+6x^1/3 part you just differentiate separately as normal so you have 6x + 2x^-2/3 + (insert derivate you get after you do the quotient rule)

Remember quitoent rule is (u’v+uv’)/v^2
This is a core 1 paper lol, isn't there an easier way?
13. https://www.derivative-calculator.net

I use this website to check my answers or help me when I’m stuck. You can use it to show steps too.
14. (Original post by HiggsBoson)
This is a core 1 paper lol, isn't there an easier way?
I’m not sure, this is how I’d instinctively do it. It would give you the right answer and it takes 30 seconds max. Do you know how to do the quotient rule? If not I’d say it is very helpful to learn (as well as the product rule but let’s not get into that just know)

So for the quotient rule you have the formula (u’v+uv’)/v^2

Take u to be the first expression in your bracket, the top part of the fraction.
So u= 2x^3 - 7
Then take v to be the bottom expression of your fraction. So v=3rootx

Now find the their derivates, u’ and v’
u’= 6x^2 and v’=3/2x^-1/2

Now put them into the formula (u’ v+uv’)/v^2 and that’s the derivate of your fraction from the brackets
15. (Original post by HiggsBoson)
7/6x^3/2 is (7/6) * x^(-3/2) simplified, and yeah it should be 6x lol
No it isn't. There are rules for which operations to do first and you need to use brackets if you want operations to be done in a different order.
16. You don't have to use quotient rule. The latter fraction can be split into multiple terms of the form ax^n.
17. (Original post by HiggsBoson)
This is a core 1 paper lol, isn't there an easier way?
Split the fraction
18. (Original post by The Night King)
I’m not sure, this is how I’d instinctively do it. It would give you the right answer and it takes 30 seconds max. Do you know how to do the quotient rule? If not I’d say it is very helpful to learn (as well as the product rule but let’s not get into that just know)

So for the quotient rule you have the formula (u’v+uv’)/v^2

Take u to be the first expression in your bracket, the top part of the fraction.
So u= 2x^3 - 7
Then take v to be the bottom expression of your fraction. So v=3rootx

Now find the their derivates, u’ and v’
u’= 6x^2 and v’=3/2x^-1/2

Now put them into the formula (u’ v+uv’)/v^2 and that’s the derivate of your fraction from the brackets
Quotient rule is not needed for this question and makes it more complicated anyway. But as it's an AS question it's clearly not expected here.
19. (Original post by RichE)
Quotient rule is not needed for this question and makes it more complicated anyway. But as it's an AS question it's clearly not expected here.
What other simpler way is there? I’m willing to learn. I’m at uni level and I’s use the quotient rule for that
20. (Original post by The Night King)
What other simpler way is there? I’m willing to learn. I’m at uni level and I’s use the quotient rule for that
Just to separate out the powers

((2x^3-7)/(3rootx)) = (2/3)x^(5/2) - (7/3)x^(-1/2)

and then differentiate

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