find-d-n-y-dx-n-for-f-x-1-1-x-

Question Number 104783 by 175mohamed last updated on 23/Jul/20

Missing \left or extra \right

Answered by Dwaipayan Shikari last updated on 23/Jul/20

f (x) = \frac{1}{\sqrt{1 - x}}

f^{^{'}} (x) = \frac{1}{2} \frac{1}{{(1 - x)}^{\frac{3}{2}}}

f^{″} (x) = \frac{3}{4} \frac{1}{{(1 - x)}^{\frac{5}{2}}}

f^{‴} (x) = \frac{15}{8} \frac{1}{{(1 - x)}^{\frac{7}{2}}}

f^{n} (x) = \frac{\underset{n = 1}{\prod^{n}} (2 n - 1)}{2^{n}} \frac{1}{{(1 - x)}^{n + \frac{1}{2}}}

Answered by mathmax by abdo last updated on 24/Jul/20

f (x) = {(1 - x)}^{- \frac{1}{2}} \Rightarrow f^{(1)} (x) = \frac{1}{2} {(1 - x)}^{- \frac{3}{2}}

f^{(2)} (x) = - \frac{3}{2^{2}} {(1 - x)}^{- \frac{5}{2}} \Rightarrow f^{(3)} (x) = \frac{3.5}{2^{3}} {(1 - x)}^{- \frac{7}{2}}

\Rightarrow f^{(n)} (x) = \frac{{(- 1)}^{n - 1} 1.3.5 \dots . (2 n - 1)}{2^{n}} {(1 - x)}^{- \frac{1}{2} - n} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n - 1} 1.3.5 \dots . (2 n - 1)}{2^{n} {(1 - x)}^{n + \frac{1}{2}}}

Commented by bubugne last updated on 24/Jul/20

w h y {(- 1)}^{n - 1} ?

Commented by abdomathmax last updated on 24/Jul/20

the sign of f^{(n)} changes \dots .

Commented by bubugne last updated on 25/Jul/20

Does the sign of f^{(n)} change ?

f (x) = {(1 - x)}^{- \frac{1}{2}} \Rightarrow f^{(1)} (x) = (- 1) (- \frac{1}{2}) {(1 - x)}^{- \frac{3}{2}}

Missing \left or extra \right

Missing \left or extra \right

f^{(3)} (x) = + \frac{15}{8} {(1 - x)}^{- \frac{7}{2}}

f^{(n)} (x) = + \frac{\underset{k = 1}{\prod^{n}} (2 k - 1)}{2^{n}} {(1 - x)}^{- \frac{2 n + 1}{2}}

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