What is (\partial)/(\partial x)((3+x^2-3y^2)^{1/2}) ?

The answer to (\partial)/(\partial x)((3+x^2-3y^2)^{1/2}) is x/((3+x^2-3y^2)^{1/2)}

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(\partial)/(\partial x)((3+x^2-3y^2)^{1/2})

\frac{\partial}{\partial x} ((3 + x^{2} - 3 y^{2})^{\frac{1}{2}})

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

Solution steps

\frac{\partial}{\partial x} ((3 + x^{2} - 3 y^{2})^{\frac{1}{2}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (3 + x^{2} - 3 y^{2})

= \frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (3 + x^{2} - 3 y^{2})

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

= \frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \cdot 2 x

Simplify

\frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \cdot 2 x : \frac{x}{( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}}

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

2 x (\frac{d y}{d x}) - 10 y = x^{- 6}

x^{2} e^{3 x}

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = b, y^{^{'}} (0) = 8

cot^{2} (sin (x))

\int ln (x) x^{'} d x

What is (\partial)/(\partial x)((3+x^2-3y^2)^{1/2}) ?
The answer to (\partial)/(\partial x)((3+x^2-3y^2)^{1/2}) is x/((3+x^2-3y^2)^{1/2)}