What is (\partial}{\partial x}(\frac{x^2-y^2)/2) ?

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

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

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

x

Solution steps

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

\frac{d}{d x} (\frac{9}{4 ( 1 + x ) ^{\frac{5}{2}}})

\int \frac{x}{3 - x ^{4}} d x

\frac{x ^{\frac{3}{2}}}{3} - x^{\frac{1}{2}}

\frac{d}{d x} (f (x) \cdot e^{\frac{1}{f ( x )}})

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