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Math Help - Help Solving Derivatives?

  1. #1
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    Help Solving Derivatives?

    This are some problems from my practice i couldn't solve.. please help me

    1. y= Ln e^4x - 1 / e^4x + 1

    2. F(x) = (tan^2 x - x^2)^3

    3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

    4. y= Ln(cos3x)

    5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

    And one integral!
    (I need to solve this one by substitution) =/

    1. |y^3(2y^2-3)dy
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by mayur03 View Post
    This are some problems from my practice i couldn't solve.. please help me

    1. y= Ln e^4x - 1 / e^4x + 1

    2. F(x) = (tan^2 x - x^2)^3

    3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

    4. y= Ln(cos3x)

    5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

    And one integral!
    (I need to solve this one by substitution) =/

    1. |y^3(2y^2-3)dy
    you do not know anything about how to derive functions show some work that's better for you I can solve them but you will not learn try .

    2)f(x)=(\tan ^2 x-x^2)^3

    f'(x)=3(\tan ^2 x -x^2)^2 (2\tan x (\sec ^2 x) - 2x)
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  3. #3
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    Quote Originally Posted by mayur03 View Post
    This are some problems from my practice i couldn't solve.. please help me

    1. y= Ln e^4x - 1 / e^4x + 1

    is this supposed to be \ln\left(\frac{e^{4x}-1}{e^{4x}+1}\right) or something else ? ... please use the correct grouping symbols.

    2. F(x) = (tan^2 x - x^2)^3

    use the chain rule ... twice.

    3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

    implicit derivative ... is that one term \ln(x^2y) ?

    4. y= Ln(cos3x)

    derivative of \ln(u) = \frac{u'}{u}

    5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

    And one integral!
    (I need to solve this one by substitution) =/

    1. |y^3(2y^2-3)dy

    no substitution is necessary ... distribute the y^3 , then integrate.
    ...
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