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

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

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

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

x

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (y^{2}) = 0

= \frac{1}{2} (2 x - 0)

Simplify

\frac{1}{2} (2 x - 0) : x

= x

\frac{d}{d t} (2 sin^{2} (t))

x \to 0 lim (x^{2} + 2 x)

\int x^{- 0.5} d x

\frac{d}{d x} (0.0225 x e^{- 0.0467 x})

\int \frac{1}{x ^{2} + 10 x + 24} d x

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