What is the derivative of y=arcsin(3/(x^2)) ?

The derivative of y=arcsin(3/(x^2)) is -(6sqrt(x^4))/(x^3sqrt(x^4-9))

What is the first derivative of y=arcsin(3/(x^2)) ?

The first derivative of y=arcsin(3/(x^2)) is -(6sqrt(x^4))/(x^3sqrt(x^4-9))

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derivative of

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

- \frac{6 x ^{4}}{x ^{3} x ^{4} - 9}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{3}{x ^{2}}) = - \frac{6}{x ^{3}}

= \frac{1}{1 - ( \frac{3}{x ^{2}} ) ^{2}} (- \frac{6}{x ^{3}})

Simplify

\frac{1}{1 - ( \frac{3}{x ^{2}} ) ^{2}} (- \frac{6}{x ^{3}}) : - \frac{6 x ^{4}}{x ^{3} x ^{4} - 9}

= - \frac{6 x ^{4}}{x ^{3} x ^{4} - 9}

\frac{d}{d t} (sin^{3} (t))

\frac{d}{d x} (3 x^{2} cos (6 x))

s (t) = 4 2 t^{2} - 4 t + 7

\int (2 x - 5 x^{2}) (2 - 10 x) d x

\int \frac{x + 1}{2 x ^{2} - 6 x + 4} d x

What is the derivative of y=arcsin(3/(x^2)) ?
The derivative of y=arcsin(3/(x^2)) is -(6sqrt(x^4))/(x^3sqrt(x^4-9))
What is the first derivative of y=arcsin(3/(x^2)) ?
The first derivative of y=arcsin(3/(x^2)) is -(6sqrt(x^4))/(x^3sqrt(x^4-9))