What is (\partial)/(\partial x)((0.5x^2+0.5y^2)^{1/2}) ?

The answer to (\partial)/(\partial x)((0.5x^2+0.5y^2)^{1/2}) is (0.5x)/((0.5x^2+0.5y^2)^{1/2)}

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(\partial)/(\partial x)((0.5x^2+0.5y^2)^{1/2})

\frac{\partial}{\partial x} ((0.5 x^{2} + 0.5 y^{2})^{\frac{1}{2}})

\frac{0.5 x}{( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial x} ((0.5 x^{2} + 0.5 y^{2})^{\frac{1}{2}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 ( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (0.5 x^{2} + 0.5 y^{2})

= \frac{1}{2 ( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (0.5 x^{2} + 0.5 y^{2})

\frac{\partial}{\partial x} (0.5 x^{2} + 0.5 y^{2}) = x

= \frac{1}{2 ( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}} x

Simplify

\frac{1}{2 ( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}} x : \frac{0.5 x}{( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}}

= \frac{0.5 x}{( 0.5 x ^{2} + 0.5 y ^{2} ) ^{\frac{1}{2}}}

\int_{0}^{\frac{π}{2}} cos^{5} (1) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 6} e^{2 t}

\frac{\partial}{\partial y} (x e^{x^{2} y})

\frac{d}{d x} (sin (x - 1))

x \to 2 lim (2 x)

What is (\partial)/(\partial x)((0.5x^2+0.5y^2)^{1/2}) ?
The answer to (\partial)/(\partial x)((0.5x^2+0.5y^2)^{1/2}) is (0.5x)/((0.5x^2+0.5y^2)^{1/2)}