# How do you find #f^6(0)# where #f(x)=arctanx/x#?

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To find ( f^6(0) ) where ( f(x) = \frac{\arctan(x)}{x} ), you can start by finding the derivatives of ( f(x) ) up to the sixth derivative and then evaluating the sixth derivative at ( x = 0 ).

The derivatives of ( f(x) ) are:

- ( f'(x) = \frac{1}{1 + x^2} - \frac{\arctan(x)}{x^2} )
- ( f''(x) = -\frac{2x}{(1 + x^2)^2} + \frac{2\arctan(x)}{x^3} - \frac{2}{x(1 + x^2)} )
- ( f'''(x) = \frac{2(3x^2 - 1)}{(1 + x^2)^3} - \frac{6\arctan(x)}{x^4} + \frac{6x}{(1 + x^2)^2} - \frac{6}{x^2(1 + x^2)} )
- ( f^{(4)}(x) = -\frac{24x(x^4 - 10x^2 + 5)}{(1 + x^2)^4} + \frac{24\arctan(x)}{x^5} - \frac{24(3x^2 - 1)}{x(1 + x^2)^3} + \frac{24}{x^3(1 + x^2)} )
- ( f^{(5)}(x) = \frac{24(5x^4 - 50x^2 + 15)}{(1 + x^2)^5} - \frac{120\arctan(x)}{x^6} + \frac{120x(x^4 - 10x^2 + 5)}{(1 + x^2)^4} - \frac{120(3x^2 - 1)}{x^2(1 + x^2)^3} + \frac{120}{x^4(1 + x^2)} )
- ( f^{(6)}(x) = -\frac{720x(5x^6 - 105x^4 + 315x^2 - 63)}{(1 + x^2)^6} + \frac{720\arctan(x)}{x^7} - \frac{720(5x^4 - 50x^2 + 15)}{x(1 + x^2)^5} + \frac{720x(x^4 - 10x^2 + 5)}{x^3(1 + x^2)^4} - \frac{720(3x^2 - 1)}{x^5(1 + x^2)^3} + \frac{720}{x^6(1 + x^2)} )

Now, evaluate ( f^{(6)}(0) ) by substituting ( x = 0 ) into the expression for the sixth derivative:

[ f^{(6)}(0) = -\frac{720 \cdot 0(5 \cdot 0^6 - 105 \cdot 0^4 + 315 \cdot 0^2 - 63)}{(1 + 0^2)^6} + \frac{720\arctan(0)}{0^7} - \frac{720(5 \cdot 0^4 - 50 \cdot 0^2 + 15)}{0(1 + 0^2)^5} + \frac{720 \cdot 0(0^4 - 10 \cdot 0^2 + 5)}{0^3(1 + 0^2)^4} - \frac{720(3 \cdot 0^2 - 1)}{0^5(1 + 0^2)^3} + \frac{720}{0^6(1 + 0^2)} ]

Simplifying this expression yields the value of ( f^{(6)}(0) ).

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