What is (\partial)/(\partial y)(x-sqrt(y^2+z^2)) ?

The answer to (\partial)/(\partial y)(x-sqrt(y^2+z^2)) is -y/(sqrt(y^2+z^2))

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(\partial)/(\partial y)(x-sqrt(y^2+z^2))

\frac{\partial}{\partial y} (x - y^{2} + z^{2})

- \frac{y}{y ^{2} + z ^{2}}

Solution steps

\frac{\partial}{\partial y} (x - y^{2} + z^{2})

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x) - \frac{\partial}{\partial y} (y^{2} + z^{2})

\frac{\partial}{\partial y} (x) = 0

\frac{\partial}{\partial y} (y^{2} + z^{2}) = \frac{y}{y ^{2} + z ^{2}}

= 0 - \frac{y}{y ^{2} + z ^{2}}

Simplify

= - \frac{y}{y ^{2} + z ^{2}}

\int_{2}^{4} 16 - x^{2} d x

6 e^{2 t}

\frac{d}{d x} (sin^{2} (cos (x)))

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\int \frac{z}{z + 1} d z

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