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C3 edexcel textbook chapter 8 differentiation Watch

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    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
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    (Original post by rainbowc4k3)
    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
    What have you tried so far?
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    (Original post by rainbowc4k3)
    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
    You know that,

    (uv)' = u'v + uv'

    Can you make a start from there?
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    (Original post by NotNotBatman)
    What have you tried so far?
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
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    (Original post by _gcx)
    You know that,

    (uv)' = u'v + uv'

    Can you make a start from there?
    I made a start, but I'm confused even though I've seen the MS
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    (Original post by rainbowc4k3)
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
    Expanding the brackets again,

    (1+3x)^4(15x+(3x+1)) = 15x(1+3x)^4 + (3x+1)(3x+1)^4 = 15x(1+3x)^4 + (3x+1)^5

    Is it clearer now?
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    (Original post by rainbowc4k3)
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
    Factorise out (1+3x)^4.

     \frac{dy}{dx} = 15x(1+3x)^4 + (1+3x)^5  = 15x(1+3x)^4 + (1+3x)^4(1+3x) because of the power rule.
    Now factorise (1+3x)^4 and consider what each term is multiplied by.

    or it might be easier to see if we let (1+3x)^4 = a, then dy/dx = 15xa^4+a^5
    Could you factorise that and then substitute a in terms of (1+3x)^4 at the end?
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    Oh wowwwwww...
    I finally get it, thank you guys so much for the help it's literally so easy now that I see it.
    Thanks guys much appreciated.
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    y = f(x)g(x), then [f(x)g(x)]’ = f’(x) g(x)+ f(x) g’(x)
    for x(1+3x)^5

    take f(x) as x

    take g(x) as (1+3x)^5

    then

    f'(x) = 1
    g'(x) = 15x(1+3x)^4

    then just substitute into rule.
 
 
 
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