What is the derivative of f(x)=arcsin(5/(x^2)) ?

The derivative of f(x)=arcsin(5/(x^2)) is -(10sqrt(x^4))/(x^3sqrt(x^4-25))

What is the first derivative of f(x)=arcsin(5/(x^2)) ?

The first derivative of f(x)=arcsin(5/(x^2)) is -(10sqrt(x^4))/(x^3sqrt(x^4-25))

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derivative of

f (x) = arcsin (\frac{5}{x ^{2}})

- \frac{10 x ^{4}}{x ^{3} x ^{4} - 25}

Solution steps

\frac{d}{d x} (arcsin (\frac{5}{x ^{2}}))

Apply the chain rule:

\frac{1}{1 - ( \frac{5}{x ^{2}} ) ^{2}} \frac{d}{d x} (\frac{5}{x ^{2}})

= \frac{1}{1 - ( \frac{5}{x ^{2}} ) ^{2}} \frac{d}{d x} (\frac{5}{x ^{2}})

\frac{d}{d x} (\frac{5}{x ^{2}}) = - \frac{10}{x ^{3}}

= \frac{1}{1 - ( \frac{5}{x ^{2}} ) ^{2}} (- \frac{10}{x ^{3}})

Simplify

\frac{1}{1 - ( \frac{5}{x ^{2}} ) ^{2}} (- \frac{10}{x ^{3}}) : - \frac{10 x ^{4}}{x ^{3} x ^{4} - 25}

= - \frac{10 x ^{4}}{x ^{3} x ^{4} - 25}

x \to 2 lim (3 π)

\frac{1}{s \cdot ( s + 1 )}

(t)^{^{'}}

n = 1 \sum \infty \frac{n + 4}{n ^{2}}

y^{^{''}} + 16 y = 0

What is the derivative of f(x)=arcsin(5/(x^2)) ?
The derivative of f(x)=arcsin(5/(x^2)) is -(10sqrt(x^4))/(x^3sqrt(x^4-25))
What is the first derivative of f(x)=arcsin(5/(x^2)) ?
The first derivative of f(x)=arcsin(5/(x^2)) is -(10sqrt(x^4))/(x^3sqrt(x^4-25))