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

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

Commented by malwaan last updated on 25/Jul/20

$$canyoupostthestepsplease$$$$thankyousirshikari$$

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

$$\frac{1.3.5.7...(2n-1)}{2^n}.\frac{2^n{x}^n}{{(1-x^2)}^{n+\frac{1}{2}}}=\frac{1.3.5.7..(2n-1)}{{(1-x^2)}^{n+\frac{1}{2}}}.x^n$$

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

$$\text{Prime causes double exponent: use braces to clarify}$$$$f^{{}^{″}}\left(x\right)=(-1)\frac{1}{2}(-\frac{3}{2})\frac{2x.2x}{{(1-x^2)}^{\frac{5}{2}}}$$$$So$$$$f^n\left(x\right)=\frac{1}{2}.\frac{3}{2}.\frac{5}{2}....n\frac{2^n.x^n}{{(1-x^2)}^{n+\frac{1}{2}}}=(1.3.5....(2n-1))\frac{x^n}{{(1-x^2)}^{n+\frac{1}{2}}}$$

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