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    Hi,

    Can some please help me with this question. I've been working on it for ages but just don't understand!

    I need to differentiate the following equation:

    Y = L + M e^(-xt)

    L = 40
    M = 10
    X = 3
    t = 4
    Is it something like dy/dx = 0 + -XMe^(-xt) or am I completely wrong?
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    (Original post by YYY)
    Hi,

    Can some please help me with this question. I've been working on it for ages but just don't understand!

    I need to differentiate the following equation:

    Y = L + M e^(-xt)

    L = 40
    M = 10
    X = 3
    t = 4
    Is it something like dy/dx = 0 + -XMe^(-xt) or am I completely wrong?
    Are there any variables in this equation? If so, which are they?

    If it's x, then it'd be dy/dx = -tMe-xt.

    If it's t, then it'd be dy/dt = -xMe-xt.

    I assume you'd then put your values in to find the value of dy/dx or dy/dt at this point.
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    (Original post by Gaiaphage)
    Are there any variables in this equation? If so, which are they?

    If it's x, then it'd be dy/dx = -tMe-xt.

    If it's t, then it'd be dy/dt = -xMe-xt.

    I assume you'd then put your values in to find the value of dy/dx or dy/dt at this point.
    Thanks for your reply. The variables are t. So I'm right but just need to put in dy/dt, not dy/dx?

    Is the L equal to 0 when differentiated?
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    (Original post by YYY)
    Thanks for your reply. The variables are t. So I'm right but just need to put in dy/dt, not dy/dx?

    Is the L equal to 0 when differentiated?
    Yeah, if t is a variable and x isn't then you'd do dy/dt. That gives dy/dt = -xMe-xt and you can put your values in from there.

    As L is a constant then yes it goes to 0 (you can think of that by thinking of the gradient of a line y = 3, the gradient would be 0)
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    (Original post by Gaiaphage)
    Yeah, if t is a variable and x isn't then you'd do dy/dt. That gives dy/dt = -xMe-xt and you can put your values in from there.

    As L is a constant then yes it goes to 0 (you can think of that by thinking of the gradient of a line y = 3, the gradient would be 0)
    Thanks for your help
 
 
 
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