What is the derivative of $arctan (\frac{8}{x^{2}})$ $arctan (8/x^2)$ ?

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Answer 1 · 2016-04-01T11:51:03

$\frac{- 16 x}{x^{4} + 64}$ $(-16x)/(x^4+64)$

$y = arctan u, u = \frac{8}{x^{2}}$ $y=arctanu, u = 8/x^2$

$\frac{d y}{d u} = \frac{1}{1 + u^{2}}, \frac{d u}{d x} = - \frac{16}{x^{3}}$ $dy/(du) = 1/(1+u^2), (du)/dx = -16/x^3$

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

$= \frac{1}{1 + {(\frac{8}{x^{2}})}^{2}} \cdot - \frac{16}{x^{3}}$ $= 1/(1+(8/x^2)^2)*-16/x^3$

$= - \frac{16}{x^{3} + \frac{64}{x}} = \frac{- 16 x}{x^{4} + 64}$ $= -16/(x^3+64/x) = (-16x)/(x^4+64)$

What is the derivative of arctan(8x2)arctan (8/x^2)?