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    Hi I am stuck on this question PLEASE HELP!!
    Find the gradient at the point x=-1 on the curve: y= x^2(2x^11+6x^4)
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    Name:  IMG_8701.jpg
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    I have done this so far not sure it is it right
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    Times out the bracket then differentiate.

    Then sub In x=-1
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    (Original post by avalerion)
    Times out the bracket then differentiate.

    Then sub In x=-1
    Thank you I got 4 as the answer
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    x^m\cdot x^n \neq x^{mn}
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    I think you need to differentiate
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    (Original post by laila_a1)
    Name:  IMG_8701.jpg
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    I have done this so far not sure it is it right
    You are right until you made dy/dx = -1
    You need to sub in -1
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    (Original post by laila_a1)
    Thank you I got 4 as the answer
    Great! I like a nice whole number
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    (Original post by avalerion)
    You are right until you made dy/dx = -1
    No they aren't.

    (Original post by laila_a1)
    Thank you I got 4 as the answer
    That's wrong.
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    (Original post by alow)
    No they aren't.



    That's wrong.
    How?
    Wait is it -44+-48 so the answer is-92 but that's a weird answer did the gradient
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    (Original post by laila_a1)
    How?
    You expanded the bracket wrong. Look at my first post.
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    Check your powers when you expanded out the bracket.
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    (Original post by laila_a1)
    How?
    Alow has pointed out a very simple mistake you've made: that multiplying x^2 and x^11 does not give x^22. Instead the indices add.

    This could perhaps make your life easier to finding an answer.
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    (Original post by benjaminfinch)
    Alow has pointed out a very simple mistake you've made: that multiplying x^2 and x^11 does not give x^22. Instead the indices add.

    This could perhaps make your life easier to finding an answer.
    Name:  image.jpg
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Size:  490.8 KB So would this be right
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    (Original post by laila_a1)
    Name:  image.jpg
Views: 35
Size:  490.8 KB So would this be right
    Yes. Well done.
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    (Original post by alow)
    Yes. Well done.
    Thank you, sorry I didn't understand you before
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    yeah, that's perfect well done
 
 
 
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