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    • Thread Starter
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    could someone differentiate
    3(5-2x)^5

    i have an answer but need to check it thnks.
    i need to find dx/dy instead of dx/dy.

    do you just put it over 1
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    Yes, just 1 over dy/dx.
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    -30(5-2x)^4 = dy/dx i think
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    (Original post by Rbates1)
    -30(5-2x)^4 = dy/dx i think
    yeh i got this too. Do i just put 1 over it to find dx/dy
    • Thread Starter
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    it then says differentiate (still dx/dy)

    1 over 3x-5
    is it just 3? doesnt seem right lol
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    \frac{1}{3x-5}=(3x-5)^{-1}
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    its  \frac{1}{3x-5}

    so that equal (3x-5)^(-1)
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    dy/dx = 2 over -3(3x-5) ??
    could someone please check this

    ps how can i write in the maths format
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    (Original post by Hrov)
    ps how can i write in the maths format
    LaTeX

    I have it bookmarked because I can't be arsed memorizing some of the weirder arguments.
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    ok thanks

    i know its the quotient rule but i want to make sure im doing it write, what is dy/dx of

    {y=}\frac{sin3x}{x}

    i got
    {y=}\frac{sin3x - x3cos3x}{x^2}

    but dont have a clue of its right
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    (Original post by Hrov)
    ok thanks

    i know its the quotient rule but i want to make sure im doing it write, what is dy/dx of

    {y=}\frac{sin3x}{x}

    i got
    {y=}\frac{sin3x - x3cos3x}{x^2}

    but dont have a clue of its right
    You have messed up the signs in the numerator of your answer.

    You don't have to use quotient rule here if you don't want to.

    y=x^{-1}sin{(3x)}

    Spoiler:
    Show
    \frac{dy}{dx}=\frac{3x\cos{(3x)}-\sin{(3x)}}{x^2}
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    (Original post by Mr M)
    You have messed up the signs in the numerator of your answer.

    You don't have to use quotient rule here if you don't want to.

    y=x^{-1}sin{(3x)}

    Spoiler:
    Show
    \frac{dy}{dx}=\frac{3x\cos{(3x)}-\sin{(3x)}}{x^2}
    thanks i got it in the end

    would 7cot5x = y

    differentiate to
    -35cosec5x
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    no
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    (Original post by D-Day)
    no
    oh well what is the correct answer, ive done it twice and get that answer, must be going wrong somewhere
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    The derivative of cotangent is -csc^2(x)
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    ok
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    (Original post by D-Day)
    The derivative of cotangent is -csc^2(x)
    -175cosec^2 25x ???
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    -35csc^2(5x)

    Chain rule.

    edit: What Mr M said.
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    (Original post by D-Day)
    -35csc^2(5x)

    Chain rule.

    edit: What Mr M said.
    Lol: I said nothing!

    I decided to remove it as it sounded a bit rude.
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    (Original post by D-Day)
    -35csc^2(5x)

    Chain rule.

    edit: What Mr M said.
    so i just missed the ^2 off???
 
 
 
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