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Question Number 104919 by mathmax by abdo last updated on 24/Jul/20

let f(x) =(1/(√(1−x^2 ))) finf f^((n)) (x)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{finf}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right) \\ $$

Commented by malwaan last updated on 25/Jul/20

can you post the steps please thank you sir shikari

$${can}\:{you}\:{post}\:{the}\:{steps}\:{please} \\ $$$${thank}\:{you}\:{sir}\:{shikari} \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jul/20

((1.3.5.7...(2n−1))/2^n ).((2^n x^n )/((1−x^2 )^(n+(1/2)) ))=((1.3.5.7..(2n−1))/((1−x^2 )^(n+(1/2)) )).x^n

$$\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}...\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} }.\frac{\mathrm{2}^{{n}} {x}^{{n}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}..\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} }.{x}^{{n}} \\ $$

Commented by Dwaipayan Shikari last updated on 25/Jul/20

f^ ′(x)=(−1)(−(1/2))((2x)/((1−x^2 )^(3/2) )) f^(′′) (x)=(−1)(1/2)(−(3/2))((2x.2x)/((1−x^2 )^(5/2) )) So f^n (x)=(1/2).(3/2).(5/2)....n((2^n .x^n )/((1−x^2 )^(n+(1/2)) ))=(1.3.5....(2n−1))(x^n /((1−x^2 )^(n+(1/2)) ))

$${f}^{\:} '\left({x}\right)=\left(−\mathrm{1}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${f}^{''} \left({x}\right)=\left(−\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)\frac{\mathrm{2}{x}.\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$$${So} \\ $$$${f}^{{n}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{2}}....{n}\frac{\mathrm{2}^{{n}} .{x}^{{n}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} }=\left(\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}−\mathrm{1}\right)\right)\frac{{x}^{{n}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$

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