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Math Help - Differentiate Y = 5 ^ (-1/x)

  1. #1
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    Differentiate Y = 5 ^ (-1/x)

    The problem is as follows: Differentiate Y = 5^(-1/x)

    My solution:

    since d/dx(a^x) = a^x * ln(a)
    I got dy/dx to be 5^(-1/x)ln5.

    this is incorrect, the answer should be [5^(-1/x)ln5] / x^2

    Thus, my question is where does that x ^ 2 denominator come from?
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  2. #2
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    Re: Differentiate Y = 5 ^ (-1/x)

    Quote Originally Posted by iAmKrizzle View Post
    The problem is as follows: Differentiate Y = 5^(-1/x)
    this is incorrect, the answer should be [5^(-1/x)ln5] / x^2
    Implicitly differentiate \ln(y)=\frac{\ln(5)}{x} and solve for y'.
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    Re: Differentiate Y = 5 ^ (-1/x)

    While the implicit differentiation seems easier, this problem is in a section before implicit differentiation is covered. This problem appears in a section on the chain rule, how would I solve it without using implicit differentiation?
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    Re: Differentiate Y = 5 ^ (-1/x)

    Quote Originally Posted by iAmKrizzle View Post
    While the implicit differentiation seems easier, this problem is in a section before implicit differentiation is covered. This problem appears in a section on the chain rule, how would I solve it without using implicit differentiation?
    Well frankly there is really no other way to do this sort of problem.
    The solution is based on this theorem:
    If f(x) has a derivative and c>0 then if y=c^{f(x)} then y'=f'(x)c^{f(x)}\ln(c).
    That theorem is proved by implicit differentiation.
    Last edited by Plato; October 7th 2011 at 02:38 PM.
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    Re: Differentiate Y = 5 ^ (-1/x)

    Understood. Thank you I will apply this principle.
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    Re: Differentiate Y = 5 ^ (-1/x)

    Quote Originally Posted by Plato View Post
    Well frankly there is really no other way to do this sort of problem.
    The solution is based on this theorem:
    If f(x) has a derivative and c>0 then if y=c^{f(x)} then y'=f'(x)c^{f(x)}\ln(c).
    That theorem is proved by implicit differentiation.
    It appears the OP is expected to know how to change the base of an exponential.

    5^{-1/x} = e^{\log \left(5^{-1/x}\right)} = e^{\frac{-1}{x} \log(5)}\right)} = e^{\frac{k}{x}} where k = -\log(5).

    After differentiating, the answer ought to be re-written in base 5 ....
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