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    Differentiate s = 4e^3t - e^-2.5t with respect to t (what does with respect to mean?)


    dy/dx (ds/de?) = 4e^2t * (3t) -e^-3.5t * (-2.5t) = 12te^t + 2.5te^-3.5t

    That is what I got. It looks completely wrong. Any help please?
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    (Original post by rm2)
    Differentiate s = 4e^3t - e^-2.5t with respect to t (what does with respect to mean?)


    dy/dx (ds/de?) = 4e^2t * (3t) -e^-3.5t * (-2.5t) = 12te^t + 2.5te^-3.5t

    That is what I got. It looks completely wrong. Any help please?
    You can't have "dy/dx" because there is no y and no x in your function definition!

    Typically you write one (dependent) variable - usually y - as a function of another (independent) variable - usually x. You then differentiate y with respect to x, which gives you the quantity you are familiar with seeing as dy/dx.

    In this case your independent variable is called t and your dependent variable is called s. You need to be calculating ds/dt which is the rate of change of s with respect to t
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    (Original post by rm2)
    Differentiate s = 4e^3t - e^-2.5t with respect to t (what does with respect to mean?)


    dy/dx (ds/de?) = 4e^2t * (3t) -e^-3.5t * (-2.5t) = 12te^t + 2.5te^-3.5t

    That is what I got. It looks completely wrong. Any help please?
    With respect to t means you will be finding ds/dt

    Have you done any work on differentiating e^x
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    Ah, yeah I thought it might be ds/dt. But other than that, I've hit a wall. Is my answer wrong?
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    (Original post by rm2)
    Ah, yeah I thought it might be ds/dt. But other than that, I've hit a wall. Is my answer wrong?
    You've gone horribly wrong because you've changed the exponents unnecessarily!

    Do you know how to work out dy/dx if y = e^{3x} ?
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    (Original post by davros)
    You've gone horribly wrong because you've changed the exponents unnecessarily!

    Do you know how to work out dy/dx if y = e^{3x} ?
    No. This is the first time I've come across differentiation like this. Usually, its just the basic nx^n-1
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    (Original post by rm2)
    No. This is the first time I've come across differentiation like this. Usually, its just the basic nx^n-1
    OK, the basic rule that you need to know is that if y = e^{kx} where k is a constant, then \dfrac{dy}{dx} = ke^{kx}

    Just replace y with s and x with t and follow the same logic.

    Are you learning this independently - it seems unusual that you would be asked to solve a problem like this if you haven't been taught the rule before?
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    We actually got taugh the thing in 1 day and didn't really venture into the harder stuff (ie this). I looked back on my notes and it turns out I wrote this formula, but its incorrect because it says if dy/dx = e^kx, then y = ke^kx so its the wrong way around.. which is why I didn't focus on that formula (I actually tried it and obbiously backfired)

    Anyway, I got ds/dt = 12e^3t + 2.5e^-2.5t ?
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    (Original post by rm2)
    We actually got taugh the thing in 1 day and didn't really venture into the harder stuff (ie this). I looked back on my notes and it turns out I wrote this formula, but its incorrect because it says if dy/dx = e^kx, then y = ke^kx so its the wrong way around.. which is why I didn't focus on that formula (I actually tried it and obbiously backfired)

    Anyway, I got ds/dt = 12e^3t + 2.5e^-2.5t ?
    This is now correct
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    (Original post by TenOfThem)
    This is now correct
    Thank you.

    I have another question that may have an incorrect formula:

    y = a sin nx - b cos nx
    dy/dx = na cos nx + nb sin nx

    y =4sin2O - 3cos4O w.r.t. O (O is theta)
    dy/dO = 8cos2O + 12sin4O

    ?
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    (Original post by rm2)
    Thank you.

    I have another question that may have an incorrect formula:

    y = a sin nx - b cos nx
    dy/dx = na cos nx + nb sin nx

    y =4sin2O - 3cos4O w.r.t. O (O is theta)
    dy/dO = 8cos2O + 12sin4O

    ?
    That is fine
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    (Original post by TenOfThem)
    That is fine
    Great, thanks!
 
 
 
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