What is (\partial)/(\partial x)(x^2-3y^2+7) ?

The answer to (\partial)/(\partial x)(x^2-3y^2+7) is 2x

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(\partial)/(\partial x)(x^2-3y^2+7)

\frac{\partial}{\partial x} (x^{2} - 3 y^{2} + 7)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} - 3 y^{2} + 7)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (3 y^{2}) + \frac{\partial}{\partial x} (7)

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (7) = 0

= 2 x - 0 + 0

Simplify

= 2 x

y^{^{'}} + y = 5

\frac{70}{x}

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

f (x) = ln (x + 1 + x^{2})

x^{3} e^{- 3 x}

What is (\partial)/(\partial x)(x^2-3y^2+7) ?
The answer to (\partial)/(\partial x)(x^2-3y^2+7) is 2x