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Thread: prove by MI that

  1. #1
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    prove by MI that

    please help with this question..

    First, it is given that f(x) is a differentiable function.
    and
    $\displaystyle
    \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)
    $

    Then, I have to prove by MI that
    $\displaystyle
    \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)
    $

    I don' t understand how to prove it . please help~
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  2. #2
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    Quote Originally Posted by ling_c_0202
    please help with this question..

    First, it is given that f(x) is a differentiable function.
    and
    $\displaystyle
    \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)
    $

    Then, I have to prove by MI that
    $\displaystyle
    \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)
    $

    I don' t understand how to prove it . please help~
    Do this like this.
    Proof (by induction) that,
    $\displaystyle [f_1(x)\cdot f_2(x)\cdot...\cdot f_n(x)]'$=$\displaystyle f'_1(x)f_2(x)\cdot...\cdot f_n(x)+$$\displaystyle f_1(x)\cdot f'_2(x)\cdot ...\cdot f_n(x)+$$\displaystyle f_1(x)f_2(x)\cdot ...\cdot f'_n(x)$
    Then,
    $\displaystyle [f(x)]^n=\underbrace{f(x)\cdot f(x)\cdot...\cdot f(x)}_{n}$
    Then by the theorem before the derivative is,
    $\displaystyle f'(x)f(x)\cdot ...\cdot f(x)+f(x)f'(x)\cdot...\cdot f(x)+$$\displaystyle f(x)\cdot f(x)\cdot...\cdot f'(x)$ - n times. But each summand can be reduced to, $\displaystyle [f(x)]^{n-1}f'(x)$ n-times. Thus,
    $\displaystyle n[f(x)]^{n-1}f'(x)$
    ---
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ling_c_0202
    please help with this question..

    First, it is given that f(x) is a differentiable function.
    and
    $\displaystyle
    \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)
    $
    The above is the base case for the induction with $\displaystyle n=2$.

    Then, I have to prove by MI that
    $\displaystyle
    \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)
    $
    Suppose this is true for some $\displaystyle k$, then

    $\displaystyle
    \frac{d}{dx}(f(x))^k = k (f(x)) ^{k-1}f '(x)
    $

    Now (using the product rule):

    $\displaystyle
    \frac{d}{dx}(f(x))^{k+1} = \frac{d}{dx} f(x) (f(x))^k $$\displaystyle
    = f(x)\frac{d}{dx}(f(x))^k + \frac{df}{dx} (f(x))^k
    $

    So applying the supposition to the first term:

    $\displaystyle
    \frac{d}{dx}(f(x))^{k+1} $$\displaystyle
    = k f(x) (f(x))^{k-1}f'(x) + f'(x) (f(x))^k
    $$\displaystyle =(k+1) (f(x))^k f'(x)$

    Which is the required result for $\displaystyle k+1$.

    So if we assume the required result for some $\displaystyle k$, it is true for $\displaystyle k+1$, hence
    with the base case the result is proven by Mathematical Induction for all $\displaystyle n \ge 2$
    (note we could have taken $\displaystyle n=1$ as the base case - which is trivialy true -
    then the proof would apply for all $\displaystyle n \ge 1$).

    RonL
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  4. #4
    Junior Member
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    thank you

    Thank you very much!
    I found that i got a mistake when i apply the product rule... stupid me
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