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f-x-1-f-x-1-f-x-

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Question Number 2071 by Rasheed Soomro last updated on 01/Nov/15
{ [f(x)]^(−1) }′ =[ f ′(x) ]^(−1) f(x)=?
{[f(x)]−1}′=[f′(x)]−1f(x)=?
Commented by Yozzi last updated on 01/Nov/15
(d/dx)((1/(f(x))))=(1/(f^′ (x))) ((f(x)×0−f^′ (x))/((f(x))^2 ))=(1/(f^′ (x))) −(f^′ (x))^2 =(f(x))^2 (f(x))^2 +(f^′ (x))^2 =0 f(x)=±if^′ (x) 1=±((if^′ (x))/(f(x))) ∫1dx=±i∫((f^′ (x))/(f(x)))dx ±iln∣f(x)∣=x+c ln∣f(x)∣=∓(i(x+c)) f(x)=e^(∓i(x+c)) f(x)=cos(x+c)+isin(x+c) or f(x)=cos(x+c)−isin(x+c) f^′ (x)=ie^(i(x+c)) ⇒(1/(f^′ (x)))=−ie^(−i(x+c)) (1/(f(x)))=e^(−i(x+c)) ⇒(d/dx)((1/(f(x))))=−ie^(−i(x+c) =f^′ (x)
ddx(1f(x))=1f′(x)f(x)×0−f′(x)(f(x))2=1f′(x)−(f′(x))2=(f(x))2(f(x))2+(f′(x))2=0f(x)=±if′(x)1=±if′(x)f(x)∫1dx=±i∫f′(x)f(x)dx±iln∣f(x)∣=x+cln∣f(x)∣=∓(i(x+c))f(x)=e∓i(x+c)f(x)=cos(x+c)+isin(x+c)orf(x)=cos(x+c)−isin(x+c)f′(x)=iei(x+c)⇒1f′(x)=−ie−i(x+c)1f(x)=e−i(x+c)⇒ddx(1f(x))=−ie−i(x+c=f′(x)
Commented by Rasheed Soomro last updated on 01/Nov/15
Like your approach!
Likeyourapproach!

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