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Math Help - more derivs...

  1. #1
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    more derivs...

    f(x) = e^(lne^(2x^(2))+3

    f(x) = e^(ln(x)+8)

    f(z) = e^(ln(9))

    can someone show me how to do these ones, so i can see the answers...much thanks...

    mathaction
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathaction View Post

    f(x) = e^(lne^(2x^(2))+3
    Brackets, there is no way of being sure what this means as:

    1. The brackets you do have don't match

    2. Because of missing brackets there is no way of telling what the "ln" is
    being applied to.

    I will assume (probably incorrectly) that you mean:

    <br />
f(x) = e^{\ln(e^{2x^2}+3)}=e^{2x^2}+3<br />

    (since e^{\ln(g)}=g) Now use the chain rule:

    <br />
f'(x)=4x e^{2x^2}<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mathaction View Post
    f(x) = e^(ln(x)+8)
    Chain rule again:

    <br />
f'(x) = (1/x) e^{\ln(x)+8}<br />

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by mathaction View Post
    f(z) = e^(ln(9))
    The right hand side is a constant, and so the derivative is zero.

    RonL
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