Find the $n^{t h}$ derivative of log $(4 - x^{2}) (| x | < 1)$ .

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Solution

Find $n^{t h}$ derivative of $log (4 - x^{2}) (| x | < 1)$
$log (4 - x^{2}) = log {(2 + x) (2 - x)} = log (2 + x) + log (2 - x)$
Now, let $f (x) = log (2 + x)$
$f_{1} (x) = \frac{1}{(2 + x)}$
$f_{2} (x) = \frac{- 1}{{(2 + x)}^{2}}$
$f_{3} (x) = (- 1) (- 2) \frac{1}{{(2 + x)}^{3}}$
..............
$f_{n} (x) = (- 1) (- 2) . . . (- (n - 1)) \frac{1}{{(2 + x)}^{n}}$
$f_{n} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + 2)}^{n}}$
Let $g (x) = log (2 - x)$
$g_{1} (x) = \frac{(- 1)}{(2 - x)}$
$g_{2} (x) = (- 1) (+ 1) \frac{1}{{(2 - x)}^{2}}$
$g_{3} (x) = (- 1) (- 2) \frac{(- 1)}{{(2 - x)}^{3}}$
..............
$g_{n} (x) = \frac{{(- 1)}^{n - 1} (n - 1)! {(- 1)}^{n}}{{(2 - x)}^{n}}$
$= \frac{{(- 1)}^{2 n - 1} (n - 1)!}{{(2 - x)}^{n}}$
$∴ n^{t h}$ derivative of $log (4 - x^{2})$ is $\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + 2)}^{n}} + \frac{{(- 1)}^{2 n - 1} (n - 1)!}{{(2 - x)}^{n}}$

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