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differentiation confused?

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Original post by kiiten
Ahh i see - i remember this method now. Do you get x^3 ?


thats perfect :h:

now what are your thinking for the 32x2 \displaystyle \frac{32}{x^2} term?
Reply 21
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Original post by DylanJ42
thats perfect :h:

now what are your thinking for the 32x2 \displaystyle \frac{32}{x^2} term?


:smile: so that would be 32x-2 which differentiates to - 64x^-3 ?
Reply 22
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Original post by kiiten
:smile: so that would be 32x-2 which differentiates to - 64x^-3 ?


Perfect.
Original post by kiiten
:smile: so that would be 32x-2 which differentiates to - 64x^-3 ?


nicely done :biggrin:
Reply 24
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Quick question. For 8.c) is it 16 - 40/3 = 2.67
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1456690494898-1124205728.jpg
Original post by kiiten
1456690494898-1124205728.jpg


answer.png

Spoiler

Reply 27
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Original post by DylanJ42
answer.png

Spoiler



Hmm i thought there might be a x2 somewhere. So, do you do 16 - 8/3 - 8/3 ?
Original post by kiiten
Hmm i thought there might be a x2 somewhere. So, do you do 16 - 8/3 - 8/3 ?


you would indeed :biggrin:
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Original post by DylanJ42
you would indeed :biggrin:


Thank you!! - Before i was doing 16 - 40/3 - 40/3 so i kept getting a -ve answer. I see where ive gone wrong now. Thanks :smile:
Reply 30
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Thanks for all your help everyone, I appreciate it :smile:
Reply 31
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Different ques - 13.a) I got x^4 = 9 Does that mean there will be a +ve & -ve answer?

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Original post by kiiten
Different ques - 13.a) I got x^4 = 9 Does that mean there will be a +ve & -ve answer?

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it does indeed :biggrin:
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Original post by DylanJ42
it does indeed :biggrin:


Ah, thanks :yes:

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Reply 34
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Original post by DylanJ42
it does indeed :biggrin:


Wait what about the shaded area? You have to prove its 5 1/3.
I integrated using the bounds 3 and 1 but got an area of 25. I think you have to use the y =4 part but i dont know how? :s

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Original post by kiiten
Wait what about the shaded area? You have to prove its 5 1/3.
I integrated using the bounds 3 and 1 but got an area of 25. I think you have to use the y =4 part but i dont know how? :s

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When you integrate between 1 and 3 you are finding the area enclosed between the curve, the x axis, x=1 and x=3 (look at the photo, area = part surrounded in green highlighter)

so you need to find area then subtract the area of the rectangle formed under the shaded area

Also check your integration, the integral shouldn't be 25

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Reply 36
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Original post by DylanJ42
its a full solution zack, but at least it's a colourful full solution :colondollar:


Not even mad. Think this is an appropriate situation for a full solution. :yep:
Reply 37
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Original post by DylanJ42
When you integrate between 1 and 3 you are finding the area enclosed between the curve, the x axis, x=1 and x=3 (look at the photo, area = part surrounded in green highlighter)

so you need to find area then subtract the area of the rectangle formed under the shaded area

Also check your integration, the integral shouldn't be 25

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So i integrated 14 - x^2 - 9x^-2 to get 14x - 1/3x^3 + 9x^-1 is that right so far?
Original post by kiiten
So i integrated 14 - x^2 - 9x^-2 to get 14x - 1/3x^3 + 9x^-1 is that right so far?


yep, all good so far :biggrin:
Reply 39
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Original post by DylanJ42
yep, all good so far :biggrin:


Then i subbed in 3 and 1 separately then took them away to get 25? :dontknow:

So 14 x 3 - 1/3 × 3^2 + 9 × 3^-1 for the first term (3)

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