What is the derivative of g(x)=(arcsin(3x))/x ?

The derivative of g(x)=(arcsin(3x))/x is (3x-arcsin(3x)sqrt(1-9x^2))/(x^2sqrt(1-9x^2))

What is the first derivative of g(x)=(arcsin(3x))/x ?

The first derivative of g(x)=(arcsin(3x))/x is (3x-arcsin(3x)sqrt(1-9x^2))/(x^2sqrt(1-9x^2))

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derivative of

g (x) = \frac{arcsin ( 3 x )}{x}

\frac{3 x - arcsin ( 3 x ) 1 - 9 x ^{2}}{x ^{2} 1 - 9 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{arcsin ( 3 x )}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( arcsin ( 3 x ) ) x - \frac{d x}{d x} arcsin ( 3 x )}{x ^{2}}

\frac{d}{d x} (arcsin (3 x)) = \frac{3}{1 - 9 x ^{2}}

\frac{d x}{d x} = 1

= \frac{\frac{3}{1 - 9 x ^{2}} x - 1 \cdot arcsin ( 3 x )}{x ^{2}}

Simplify

\frac{\frac{3}{1 - 9 x ^{2}} x - 1 \cdot arcsin ( 3 x )}{x ^{2}} : \frac{3 x - arcsin ( 3 x ) 1 - 9 x ^{2}}{x ^{2} 1 - 9 x ^{2}}

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

x \to - \infty lim (- 3 x^{3} + x)

x \to 0 lim (- \frac{ln ( 1 + 6 x )}{x})

f (t) = \frac{t sin ( t )}{1 + t}

\int \frac{2 - x}{x} d x

\int x^{2} d x

What is the derivative of g(x)=(arcsin(3x))/x ?
The derivative of g(x)=(arcsin(3x))/x is (3x-arcsin(3x)sqrt(1-9x^2))/(x^2sqrt(1-9x^2))
What is the first derivative of g(x)=(arcsin(3x))/x ?
The first derivative of g(x)=(arcsin(3x))/x is (3x-arcsin(3x)sqrt(1-9x^2))/(x^2sqrt(1-9x^2))