What is (\partial)/(\partial y)(sqrt(1-y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(1-y^2)) is -y/(sqrt(1-y^2))

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(\partial)/(\partial y)(sqrt(1-y^2))

\frac{\partial}{\partial y} (1 - y^{2})

- \frac{y}{1 - y ^{2}}

Solution steps

\frac{\partial}{\partial y} (1 - y^{2})

Apply the chain rule:

\frac{1}{2 1 - y ^{2}} \frac{\partial}{\partial y} (1 - y^{2})

= \frac{1}{2 1 - y ^{2}} \frac{\partial}{\partial y} (1 - y^{2})

\frac{\partial}{\partial y} (1 - y^{2}) = - 2 y

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

Simplify

\frac{1}{2 1 - y ^{2}} (- 2 y) : - \frac{y}{1 - y ^{2}}

= - \frac{y}{1 - y ^{2}}

2 sin (x)

\int \frac{2 x + 1}{e ^{2 x}} d x

\frac{\partial}{\partial x} (e^{x} cos (y) + 2 x - 3 y^{2})

2 + t + \frac{1}{t}

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

What is (\partial)/(\partial y)(sqrt(1-y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(1-y^2)) is -y/(sqrt(1-y^2))