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Math Help - Finding f'(x)

  1. #1
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    Finding f'(x)

    I don't even know where to start with this:

    d/dx f(sin(x))=1/(cos(x))

    find a formula for f'(x)

    This is a review question and I've tried expanding f'(sin(x)) but nothing

    I just need a hint to how I get started on this problem

    Oh I know the answer to f'(x) is 1/(1-x^2) but I got it from trial and error is there a mathematical way
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  2. #2
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    d/dx f(sin(x)) = f'(sin(x))cos(x) = 1/cos(x), from where f'(sin(x)) = 1/cos^2(x) = 1/(1 - sin^2(x)). I think this implies that f'(x) = 1/(1-x^2), but, strictly speaking, only for |x| <= 1.
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  3. #3
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    dy/dx = 1/cos x = sec x
    dy = sec x dx
    y = ∫ sec x dx = ln|sec x + tan x| + C
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  4. #4
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    Quote Originally Posted by ppark View Post
    I don't even know where to start with this:

    d/dx f(sin(x))=1/(cos(x))
    Let u= sin(x). Then du/dx= cos(x). By the chain rule, \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}= cos(x)\frac{df}{du} so that equation becomes cos(x)\frac{df}{dx}= \frac{1}{cos(x)} which is the same as \frac{df}{du}= \frac{1}{cos^2(x)}= \frac{1}{1- sin^2(x)}= \frac{1}{1- u^2}. Replacing "u" with "x", that is \frac{df}{dx}= \frac{1}{1- x^2}.

    find a formula for f'(x)

    This is a review question and I've tried expanding f'(sin(x)) but nothing


    I just need a hint to how I get started on this problem

    Oh I know the answer to f'(x) is 1/(1-x^2) but I got it from trial and error is there a mathematical way
    I feel I should point out that "trial and error" is a perfectly valid "mathematical way" of solving a problem!
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  5. #5
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    Quote Originally Posted by johnny View Post
    dy/dx = 1/cos x = sec x
    dy = sec x dx
    y = ∫ sec x dx = ln|sec x + tan x| + C
    I think you misunderstood the question.
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