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  • <img alt="\text { Given } y=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \text { then }\left(\frac{d^5 y}{d x^5}\right)_{x=0}=" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%20%7B%20Give

\text { Given } y=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) \text { then }\left(\frac{d^5 y}{d x^5}\right)_{x=0}=

Option: 1

84


Option: 2

 48

 


Option: 3

12


Option: 4

-\frac{1}{12}


Answers (1)

best_answer

\text { Let } y=\tan ^{-1}\left\{2 x /\left(1-x^2\right)\right\}=2 \tan ^{-1} x \text {. }

then y_1=\frac{2}{1+x^2}=\frac{2}{(x-i)(x+i)}=\frac{1}{i}\left[\frac{1}{x-i}-\frac{1}{x+i}\right]

on resolving into partial fractions.Now differentiating both sides (n-1) times w.r.t. 'x' , we have

\mathrm{y}_{\mathrm{n}}=\frac{(-1)^{\mathrm{n}-1}(\mathrm{n}-1) !}{\mathrm{i}}\left[(\mathrm{x}-\mathrm{i})^{-\mathrm{n}}-(\mathrm{x}+\mathrm{i})^{-\mathrm{n}}\right]

Putting x=r \cos \phi \text { and } 1-r \sin \phi \text {, we have }

\mathrm{y}_{\mathrm{n}}=\frac{(-1)^{\mathrm{n}-1}(\mathrm{n}-1) !}{\mathrm{i}}\left[\mathrm{r}^{-\mathrm{n}}(\cos \phi-\mathrm{i} \sin \phi)^{-\mathrm{n}}-\mathrm{r}^{-\mathrm{n}}(\cos \phi+\mathrm{i} \sin \phi)^{-\mathrm{n}}\right]\begin{aligned} & =\frac{(-1)^{\mathrm{n}-1}(\mathrm{n}-1) !}{\mathrm{i}} \mathrm{r}^{-\mathrm{n}}[(\cos \mathrm{n} \phi+\sin \mathrm{n} \phi)-(\cos \mathrm{n} \phi-\mathrm{i} \sin \mathrm{n} \phi)] \\ & =2(-1)^{\mathrm{n}-1}(\mathrm{n}-1) ! \mathrm{r}^{-\mathrm{n}} \sin \mathrm{n} \phi \\ & =2(-1)^{\mathrm{n}-1}(\mathrm{n}-1) !(1 / \sin \phi)^{-\mathrm{n}} \sin \mathrm{n} \phi, \sin \operatorname{ce} \mathrm{r}=1 / \sin \phi \\ & =2(-1)^{\mathrm{n}-1}(\mathrm{n}-1) ! \sin ^{\mathrm{n}} \phi \sin \mathrm{n} \phi, \text { where } \phi=\tan ^{-1}(1 / \mathrm{x}) \end{aligned}

 

Posted by

avinash.dongre

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