What is the derivative of 6-arcsin(x/(x^2+4)) ?

The derivative of 6-arcsin(x/(x^2+4)) is -(-x^2+4)/((x^2+4)sqrt(x^4+7x^2+16))

What is the first derivative of 6-arcsin(x/(x^2+4)) ?

The first derivative of 6-arcsin(x/(x^2+4)) is -(-x^2+4)/((x^2+4)sqrt(x^4+7x^2+16))

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\frac{d}{d x} (6 - arcsin (\frac{x}{x ^{2} + 4}))

- \frac{- x ^{2} + 4}{( x ^{2} + 4 ) x ^{4} + 7 x ^{2} + 16}

Solution steps

\frac{d}{d x} (6 - arcsin (\frac{x}{x ^{2} + 4}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (6) - \frac{d}{d x} (arcsin (\frac{x}{x ^{2} + 4}))

\frac{d}{d x} (6) = 0

\frac{d}{d x} (arcsin (\frac{x}{x ^{2} + 4})) = \frac{- x ^{2} + 4}{( x ^{2} + 4 ) x ^{4} + 7 x ^{2} + 16}

= 0 - \frac{- x ^{2} + 4}{( x ^{2} + 4 ) x ^{4} + 7 x ^{2} + 16}

Simplify

= - \frac{- x ^{2} + 4}{( x ^{2} + 4 ) x ^{4} + 7 x ^{2} + 16}

\int (3 x^{2} - 4 x^{3}) d x

x y^{^{'}} + 3 y^{^{'}} - (x + 3) cos (x) + y = 0

\int (1 - y)^{3} d y

\int y cos (4 y) d y

y = \frac{( 2 x ^{2} - 1 ) ^{3}}{( 5 x - 8 ) ^{2}}

What is the derivative of 6-arcsin(x/(x^2+4)) ?
The derivative of 6-arcsin(x/(x^2+4)) is -(-x^2+4)/((x^2+4)sqrt(x^4+7x^2+16))
What is the first derivative of 6-arcsin(x/(x^2+4)) ?
The first derivative of 6-arcsin(x/(x^2+4)) is -(-x^2+4)/((x^2+4)sqrt(x^4+7x^2+16))