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

The derivative of arcsin((x^2+y^2/2)) is (2(x+y\frac{dy)/(dx))}{sqrt(4-x^4-2x^2y^2-y^4)}

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

The first derivative of arcsin((x^2+y^2/2)) is (2(x+y\frac{dy)/(dx))}{sqrt(4-x^4-2x^2y^2-y^4)}

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\frac{d}{d x} (arcsin (\frac{x ^{2} + y ^{2}}{2}))

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

Solution steps

\frac{d}{d x} (arcsin (\frac{x ^{2} + y ^{2}}{2}))

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} + y ^{2}}{2}) = x + y \frac{d y}{d x}

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

Simplify

\frac{1}{1 - ( \frac{x ^{2} + y ^{2}}{2} ) ^{2}} (x + y \frac{d y}{d x}) : \frac{2 ( x + y \frac{d y}{d x} )}{4 - x ^{4} - 2 x ^{2} y ^{2} - y ^{4}}

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

y = \frac{6 x}{x ^{2} + 1}, (1, 3)

\int \frac{1}{2 \cdot x} d x

\frac{1}{x + 3}

e^{- 3 t} cos (2 t)

\int_{0}^{2} 9 x + 37 d x

What is the derivative of arcsin((x^2+y^2/2)) ?
The derivative of arcsin((x^2+y^2/2)) is (2(x+y\frac{dy)/(dx))}{sqrt(4-x^4-2x^2y^2-y^4)}
What is the first derivative of arcsin((x^2+y^2/2)) ?
The first derivative of arcsin((x^2+y^2/2)) is (2(x+y\frac{dy)/(dx))}{sqrt(4-x^4-2x^2y^2-y^4)}