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

Question Number 2071 by Rasheed Soomro last updated on 01/Nov/15

{{[f (x)]}^{- 1}}^{'} = {[f^{'} (x)]}^{- 1}

f (x) = ?

Commented by Yozzi last updated on 01/Nov/15

\frac{d}{d x} (\frac{1}{f (x)}) = \frac{1}{f^{^{'}} (x)}

\frac{f (x) \times 0 - f^{^{'}} (x)}{{(f (x))}^{2}} = \frac{1}{f^{^{'}} (x)}

- {(f^{^{'}} (x))}^{2} = {(f (x))}^{2}

{(f (x))}^{2} + {(f^{^{'}} (x))}^{2} = 0

f (x) = \pm {i f}^{^{'}} (x)

1 = \pm \frac{{i f}^{^{'}} (x)}{f (x)}

\int 1 d x = \pm i \int \frac{f^{^{'}} (x)}{f (x)} d x

\pm i l n ∣ f (x) ∣= x + c

l n ∣ f (x) ∣= \mp (i (x + c))

f (x) = e^{\mp i (x + c)}

f (x) = c o s (x + c) + i s i n (x + c)

o r

f (x) = c o s (x + c) - i s i n (x + c)

f^{^{'}} (x) = {i e}^{i (x + c)} \Rightarrow \frac{1}{f^{^{'}} (x)} = - {i e}^{- i (x + c)}

\frac{1}{f (x)} = e^{- i (x + c)} \Rightarrow \frac{d}{d x} (\frac{1}{f (x)}) = - {i e}^{- i (x + c} = f^{^{'}} (x)

Commented by Rasheed Soomro last updated on 01/Nov/15

L i k e y o u r a p p r o a c h!