How do you find the derivative of $f (x) = \frac{arcsin (3 x)}{x}$ $f(x) =(arcsin(3x))/x$ ?

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How do you find the derivative of $f (x) = \frac{arcsin (3 x)}{x}$ $f(x) =(arcsin(3x))/x$ ?

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Answer 1 · 2018-03-14T18:41:04

The answer would be $\frac{3 \cdot \sqrt{1 - 9 x^{2}}}{x \cdot (1 - 9 x^{2})} - \frac{arcsin (3 x)}{x^{2}}$ $(3*sqrt(1-9x^2))/(x*(1-9x^2))-arcsin(3x)/x^2$

Explanation:

$f (x) = \frac{arcsin (3 x)}{x}$ $f(x)=arcsin(3x)/x$

$\frac{d}{d x} (\frac{arcsin (3 x)}{x})$ $d/dx(arcsin(3x)/x)$

$= \frac{\frac{d}{d x} (arcsin (3 x)) \cdot x - \frac{d}{d x} (x) \cdot arcsin (3 x)}{x^{2}}$ $=(d/dx(arcsin(3x))*x-d/dx(x)*arcsin(3x))/x^2$

$= \frac{\frac{\frac{d}{d x} (3 x)}{\sqrt{1 - {(3 x)}^{2}}} - arcsin (3 x)}{x^{2}}$ $=((d/dx(3x))/sqrt(1-(3x)^2)-arcsin(3x))/x^2$

$= \frac{\frac{3 x}{\sqrt{1 - 9 x^{2}}} - arcsin (3 x)}{x^{2}}$ $=((3x)/sqrt(1-9x^2)-arcsin(3x))/x^2$

$= \frac{(3 x) \sqrt{1 - 9 x^{2}}}{(x^{2}) (1 - 9 x^{2})} - \frac{arcsin (3 x)}{x^{2}}$ $=((3x)sqrt(1-9x^2))/((x^2)(1-9x^2))-arcsin(3x)/x^2$

$= \frac{3 \sqrt{1 - 9 x^{2}}}{x (1 - 9 x^{2})} - \frac{arcsin (3 x)}{x^{2}}$ $=(3sqrt(1-9x^2))/(x(1-9x^2))-arcsin(3x)/x^2$

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How do you find the derivative of $f (x) = \frac{arcsin (3 x)}{x}$ $f(x) =(arcsin(3x))/x$ ?

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