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    Hey,
    Just doing some revision before the exam on Wednesday and I came across this question, I missed out the lessons on the product rule and quotient rule but I managed to pick it up pretty quickly but I was revising the chapter in my text book and I cant do this one question:



    I tried it myself obviously but I just cant seem to get the right answer :mad:

    Thanks for any help
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    (Original post by THRASHx)
    Hey,
    Just doing some revision before the exam on Wednesday and I came across this question, I missed out the lessons on the product rule and quotient rule but I managed to pick it up pretty quickly but I was revising the chapter in my text book and I cant do this one question:



    I tried it myself obviously but I just cant seem to get the right answer :mad:

    Thanks for any help
    You need to use the chain rule in this question ....I'm not sure if you're aware of that, and then it should be fairly straight forward.... :/
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    Well, let u=x^2 and v = (2x+1)^4 and off you go...
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    From the post above, use the chain rule for V to get 8(2x+1)^3 to get dv/dx then use the product rule.
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    I got
    dy/dx= 2x^3 + 8(2x+1)^4 (2x+1)^3
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    or you could just multiply it out using the binomial theorem and differentiate term by term...
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    i like how you blurred out the rest and highlighted that question, nice feature.

    anyway, yeah use the product rule as said above and when differentiation
    (2x+1)^4
    use the chain rule.
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    where could I find the answer?
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    (Original post by xXxBaby-BooxXx)
    You need to use the chain rule in this question ....I'm not sure if you're aware of that, and then it should be fairly straight forward.... :/
    Chain and product:yes:
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    (Original post by Kolya)
    Well, let u=x^2 and v = (2x+1)^4 and off you go...
    I did that, but I ended up getting something different to the answer although im pretty sure I did it right :confused:
    Hence why I came on here.

    The answer is 2x(6x+1)(2x+1)^3 I know id have to rearrange my answer but I cant see how it gets to that, so I think ive gone wrong somewhere.
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    dy/dx = 2x[(2x + 1)^4] + x^2[4 x 2(2x + 1)^3]
    = 2x[(2x + 1)^4] + 8x^2[(2x + 1)^3]
    = 2x(2x + 1)^3((2x + 1) + 4x)
    = 2x(2x + 1)^3(6x + 1)
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    (i think)
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    (Original post by THRASHx)
    I did that, but I ended up getting something different to the answer although im pretty sure I did it right :confused:
    Hence why I came on here.

    The answer is 2x(6x+1)(2x+1)^3 I know id have to rearrange my answer but I cant see how it gets to that, so I think ive gone wrong somewhere.
    If you use chain rule and product rule as everyone said, you'll end up with:
    2x(2x+1)^4 + 8x^2(2x+1)^3
    which you can factorise to get the answer
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    (Original post by Toneh)
    If you use chain rule and product rule as everyone said, you'll end up with:
    2x(2x+1)^4 + 8x^2(2x+1)^3
    which you can factorise to get the answer
    Yeah, I tried it. Im not the kinda person who comes on here to ask people without even trying first!

    Ok looking at it I see where ive gone wrong, when I got x^2 . 8(2x+1)^3 for uv' I multiplied it wrong like an idiot.

    Thanks everyone.
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    this is a mix of product and chain rule.

    keep 1st, diff 2nd. keep 2nd, diff 1st

    check the attachment
    Attached Images
     
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    (Original post by Drederick Tatum)
    or you could just multiply it out using the binomial theorem and differentiate term by term...
    yup, sounds like a nice waste of time

    y = x^2(2x + 1)^4

    u = x^2

    v = (2x + 1)^4

    y = uv

    \frac {du}{dx} = 2x

    \frac {dv}{dx} = 8(2x + 1)^3

    \frac {dy}{dx} = 2x(2x + 1)^4 + 8x^2(2x + 1)^3 = 2x(2x + 1)^3((2x + 1) + 4x) = 2x(2x + 1)^3(6x + 1)
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    (Original post by THRASHx)
    Yeah, I tried it. Im not the kinda person who comes on here to ask people without even trying first!

    Ok looking at it I see where ive gone wrong, when I got x^2 . 8(2x+1)^3 for uv' I multiplied it wrong like an idiot.

    Thanks everyone.
    Haha, sorry I wasn't insinuating that you hadn't tried it, but was simply emphasising that my post was basically pointless as everyone had already told you what to do.
    :P
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    (Original post by Toneh)
    Haha, sorry I wasn't insinuating that you hadn't tried it, but was simply emphasising that my post was basically pointless as everyone had already told you what to do.
    :P
    Its fine at least you didnt suggest binomial expansion :gasp:
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    (Original post by Pheylan)
    yup, sounds like a nice waste of time
    It's not though, because you can just write down the answer straight away doing this. It depends what you need the answer for..
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    (Original post by Drederick Tatum)
    It's not though, because you can just write down the answer straight away doing this. It depends what you need the answer for..
    It doesnt help my differentiation revision at all though does it?
 
 
 
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