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    (Original post by insparato)
    y = e^2x + e^-x

    dy/dx = 2e2x -ex

    dy/dx = 0

    2e2x -ex = 0

    ex ( 2ex - 1) = 0

    ex = 0

    ex = 1/2

    x = 0
    x = ln (0.5) = ln1 - ln2 = -ln2

    so sup these back into the original

    y = 2 when x = 0

    y = 2 + 1/4 = 9/4

    (0,2) (ln2,9/4)
    i dont think you've worked out dy/dx correctly. should it not be 2e^2x - e^-x as opposed to 2e^2x -e^x
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    another question:
    a) find the gradient of y = 2ln x - x^2 at the point where x = 2
    how do you find dy/dx? do you have to use the quotient rule and if so, do you get the answer as (2x - 4ln x^2)/x^4?

    b) find the coordinates of the stationary point on the graph.
    i know you have to make dy/dx = 0 but dont know how to factorise it
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    also, is there a way that you can attach documents with scanned pictures in it becuase it says the document is too big to attach
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    (Original post by alison_141288)
    another question:
    a) find the gradient of y = 2ln x - x^2 at the point where x = 2
    how do you find dy/dx? do you have to use the quotient rule and if so, do you get the answer as (2x - 4ln x^2)/x^4?
    The quotient rule is for fractions, not sums/differences - to differentiate something like u + v or u - v, you just diffentiate each part in turn. (You should then find part (b) much easier, once you get (a) right.)
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    (Original post by alison_141288)
    another question:
    a) find the gradient of y = 2ln x - x^2 at the point where x = 2
    how do you find dy/dx? do you have to use the quotient rule and if so, do you get the answer as (2x - 4ln x^2)/x^4?

    b) find the coordinates of the stationary point on the graph.
    i know you have to make dy/dx = 0 but dont know how to factorise it
    quotient rule is for differentiating fractions. How is this function a fraction? :confused:

    a) dy/dx = 2/x - 2x
    At x = 2, dy/dx = 2/2 - 2(2) = 1-4 = -3

    b) Stationary points occur at dy/dx = 0
    => 2/x - 2x = 0
    => 2/x = 2x
    => 2 = 2x^2
    => x^2 = 1
    x = 1 (not -1 since ln(-1) is undefined)

    At x = 1
    y = 2ln1 - 1^2
    y = 0 - 1
    y = -1

    Therefore stationary point is (1,-1)

    I think you need a bit more practice on understanding the theory rather than doing questions.
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    how do you differentiate:
    a) ln 6x
    b) 2ln x/4
    c) ln 4/x

    is a? 1/x becuase ln6 is a constant so you just differntiate lnx. however, when you use the product rule, i get the answer of ln6 so which is the correct answer/method?
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    (Original post by alison_141288)
    how do you differentiate:
    a) ln 6x
    b) 2ln x/4
    c) ln 4/x

    is a? 1/x becuase ln6 is a constant so you just differntiate lnx. however, when you use the product rule, i get the answer of ln6 so which is the correct answer/method?
    a) There are two ways of doing this.
    Write ln6x as ln6 + lnx and differentiate as normal.
    Or differentiate ln6x using the chain rule.
    d(ln6x)/dx = 6/6x = 1/x

    b) d(2ln(x/4))/dx = 2(1/4)/(x/4) = 2/x

    c) d(ln(4/x))/dx = d(ln4-lnx)/dx = -1/x
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    (Original post by Widowmaker)
    a) There are two ways of doing this.
    Write ln6x as ln6 + lnx and differentiate as normal.
    Or differentiate ln6x using the chain rule.
    d(ln6x)/dx = 6/6x = 1/x

    b) d(2ln(x/4))/dx = 2(1/4)/(x/4) = 2/x

    c) d(ln(4/x))/dx = d(ln4-lnx)/dx = -1/x
    what's the chain rule? is it the same as the product rule
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    (Original post by alison_141288)
    what's the chain rule? is it the same as the product rule
    Does your teacher actually go through the theory in class? Or are you self teaching?

    The chain rule is fundamental to calculus!

    Look up chain rule in the C3 heinemann book and read up on it.

    Okay here's a brief summary

    CHAIN RULE
    If you have a function f(g(x) = F(x)
    F'(x) = f'(g(x)) x g'(x)

    PRODUCT RULE
    If you have a function uv
    Then d(uv)/dx = u.dv/dx + v.du/dx

    QUOTIENT RULE
    If you have a function u/v
    d(u/v)/dx = (v.du/dx - u.dv/dx)/v^2
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    (Original post by Widowmaker)
    Does your teacher actually go through the theory in class? Or are you self teaching?

    The chain rule is fundamental to calculus!

    Look up chain rule in the C3 heinemann book and read up on it.

    Okay here's a brief summary

    CHAIN RULE
    If you have a function f(g(x) = F(x)
    F'(x) = f'(g(x)) x g'(x)

    PRODUCT RULE
    If you have a function uv
    Then d(uv)/dx = u.dv/dx + v.du/dx

    QUOTIENT RULE
    If you have a function u/v
    d(u/v)/dx = (v.du/dx - u.dv/dx)/v^2
    our teacher does go through it but i missed a lesson last week and it's so much hassle trying to catch up with all the work. never heard of the chain rule and dont recognise it or understand your explanation but i miss my lesson again tomorrow and on thurs so will have to wait till next tues to ask her
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    y = e2x + e-x

    dy/dx = 2e2x - e-x

    stationary points occur when dy/dx = 0

    2e2x - e-x = 0

    ex(2ex - e-2x) = 0

    ex = 0

    no solution as corrected

    2ex - e-2x = 0

    2ex = e-2x

    2ex/e-2x = 1

    2e3x = 1

    e3x = 1/2

    3x = ln(1/2)

    x= 1/3ln(1/2)

    x = -1/3ln2

    so

    x = -1/3ln2 = ln (2)-1/3

    sub x = -1/3ln2 into y = e2x + e-x

    y = e-2/3ln2 + e1/3ln2

    y = eln 2-2/3 + eln21/3

    y = 2-2/3 + 21/3

    Dont think its wise to put that into a number.

    so (-1/3ln2, 2-2/3 + 21/3)

    Somebody kind might want to check ive done it right this time. Computers are so bad for writing maths on.If i have to use the [sup] command again im going to go kill myself.
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    ha yes not on the ball today.
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    nobody will ever know...
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    (Original post by insparato)
    y = e2x + e-x

    dy/dx = 2e2x - e-x

    stationary points occur when dy/dx = 0

    2e2x - e-x = 0

    ex(2ex - e-2x) = 0

    ex = 0

    no solution as corrected

    2ex - e-2x = 0

    2ex = e-2x

    2ex/e-2x = 1

    2e3x = 1

    e3x = 1/2

    3x = ln(1/2)

    x= 1/3ln(1/2)

    x = -1/3ln2

    so

    x = -1/3ln2 = ln (2)-1/3

    sub x = -1/3ln2 into y = e2x + e-x

    y = e-2/3ln2 + e1/3ln2

    y = eln 2-2/3 + eln21/3

    y = 2-2/3 + 21/3

    Dont think its wise to put that into a number.

    so (-1/3ln2, 2-2/3 + 21/3)

    Somebody kind might want to check ive done it right this time. Computers are so bad for writing maths on.If i have to use the [sup] command again im going to go kill myself.
    how did you get this line of working? surely it would leave it as 0 if you divide 2e^x by e^-2x
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    2e^x = e^-2x

    2e^x / e^-2x = e^-2x / e^-2x

    2e^x / e^-2x = 1
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    (Original post by alison_141288)
    how did you get this line of working? surely it would leave it as 0 if you divide 2e^x by e^-2x
    2e^x/e^-2x = 2e^x*e^2x
    = 2e^3x
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    Laws of indices my apologies for not being clearer.

    when you divide numbers with powers you take away the powers.
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    Have I read the wrong line
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    I think between us, we've answered any possible query with that line!
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    x= 1/3ln(1/2)

    x = -1/3ln2

    i get the first line but how do you get to the second line of working?
 
 
 
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