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Math Help - derivatives

  1. #1
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    derivatives

    Hey guys, I am trying to complete a take home exam that is due in the morning, and was hoping that I could get some help on a few of them I am not sure about.

    Find the derivative of the function. Assume ab,c, and k are constants.

    f(x)= e^(2x)*((x^2)+(5^x))

    f(w)= ((5w^2)+3)*e^w^2

    f(x)=1/((e^x)+1)^2

    f(x)=e^(-(x-1)^2)

    f(w)=ln(cos(w-1))

    y=2x(ln(x)+ln(2))-2x+e


    Thank you in advance and sorry if I confused you guys trying to write out the equations.
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  2. #2
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    Okay, we can actually help ya, but can't actually do your entire homework, you must tell us what you've done, at least. (Or specify where you're gettin' stuck.)
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  3. #3
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    hi,

    1.)
    f'(x)=2e^2x(x^2+5^x)+e^2x(2x+5^xlog5)

    2.)

    f'(x)=(10w)e^w^2+(5w^2+3)2w(e^w^2)

    3.)
    f'(x)=2(i/5^x+1)(-5^xlog5/5^2x)

    4.)
    f'(w)=[1/cos(w-1)]sin(w-1)

    best of luck
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  4. #4
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    hi,

    1.)
    f'(x)=2e^2x(x^2+5^x)+e^2x(2x+5^xlog5)

    2.)

    f'(x)=(10w)e^w^2+(5w^2+3)2w(e^w^2)

    3.)
    f'(x)=2(i/5^x+1)(-5^xlog5/5^2x)

    4.)
    f'(w)=[1/cos(w-1)]sin(w-1)=tan(w-1)

    best of luck
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    LaTeX isn't working
    Quote Originally Posted by Jea9 View Post
    Hey guys, I am trying to complete a take home exam that is due in the morning, and was hoping that I could get some help on a few of them I am not sure about.
    why're you just asking for help now?

    Find the derivative of the function. Assume ab,c, and k are constants.

    f(x)= e^(2x)*((x^2)+(5^x))
    do you remember the product rule?

    it says, d/dx [f(x)*g(x)] = f '(x)*g(x) + f(x) * g'(x)

    here, you have f(x) = e^{2x} and g(x) = x^2 + 5^x

    i suppose you know the derivatives of things of the form e^u, x^n and a^x (where a is a constant greater than 0). can you continue?



    f(w)= ((5w^2)+3)*e^w^2
    same story here. product rule. use f(w) = 5w^2 + 3 and g(w) = e^{w^2} in the formula i gave you above

    f(x)=1/((e^x)+1)^2
    note that 1/(e^x + 1)^2 = (e^x + 1)^{-2}

    now use the chain rule. do you remember it? it says:

    d/dx [f(g(x))] = f ' (g(x)) * g'(x)

    here, f(x) = x^{-2} and g(x) = e^x + 1

    do you see that? can you continue?

    f(x)=e^(-(x-1)^2)
    again, chain rule: f(x) = e^x and g(x) = -(x - 1)^2 in the formula above. (i hope it doesn't confuse you that i am using f(x) in my formula when you have f(x) in your problem )

    f(w)=ln(cos(w-1))
    chain rule yet again: use f(w) = ln(w) and g(w) = cos(w - 1)...note that you need the chain rule to derive cos(w - 1) as well, but it is no big deal, since the derivative of w - 1 is 1


    y=2x(ln(x)+ln(2))-2x+e
    use the product rule for the first term. differentiate the rest normally. can you continue?


    sorry if I confused you guys trying to write out the equations.
    you do better than most
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  6. #6
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    oh sorry. alright well with the first one this is what I have as the answer.

    f '(x)=2e^(2x)((x^2)+(5^x))(2x+ln(5)(5x))

    I'm not sure whether or not I applied the chain rule correctly here, what I did was take the derivative of the outer equation e^2x and then that of the inner. The ln really tripped me up. a^x is equal to ln(a) * a^x, right? The second one I worked out similarly.

    f '(w)=e^(w^2)*((5w^2)+3)(2w)(10w)
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jea9 View Post
    oh sorry. alright well with the first one this is what I have as the answer.

    f '(x)=2e^(2x)((x^2)+(5^x))(2x+ln(5)(5x))
    nope. follow the product rule. you should get a sum as the answer, so there should be two terms. and i think you mean 5^x where you typed 5x

    I'm not sure whether or not I applied the chain rule correctly here, what I did was take the derivative of the outer equation e^2x and then that of the inner.
    yeah, you messed up. see my comments above as well as my first post for the rule.

    The ln really tripped me up.
    ln trips everyone up

    the derivative of a^x is equal to ln(a) * a^x, right?
    yes

    The second one I worked out similarly.

    f '(w)=e^(w^2)*((5w^2)+3)(2w)(10w)
    you made the same mistake here. try to fix it
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  8. #8
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    Oh thank you. I see the problem. I think I mix up when to use the chain rule and when to use the product rule. Thank you. I am going to look over my other problems and see if I made that mistake with the other ones. If I have another problem I will post it. Thanks everyone.
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  9. #9
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    I just went through the exam again. It seems that same product rule vs. chain rule thing surfaced in most of the problems I did wrong. I will have to work on that. Thanks everyone for the help.
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