What is (\partial)/(\partial x)(2x^2-7y^9-4) ?

The answer to (\partial)/(\partial x)(2x^2-7y^9-4) is 4x

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(\partial)/(\partial x)(2x^2-7y^9-4)

\frac{\partial}{\partial x} (2 x^{2} - 7 y^{9} - 4)

4 x

Solution steps

\frac{\partial}{\partial x} (2 x^{2} - 7 y^{9} - 4)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x^{2}) - \frac{\partial}{\partial x} (7 y^{9}) - \frac{\partial}{\partial x} (4)

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

\frac{\partial}{\partial x} (7 y^{9}) = 0

\frac{\partial}{\partial x} (4) = 0

= 4 x - 0 - 0

Simplify

= 4 x

e^{x} sin (x) + e^{x} cos (x)

\int y y^{2} + 1 d y

x \to 3 lim (2 x + 4)

n = 0 \sum \infty 2 \cdot 0. 3^{n}

What is (\partial)/(\partial x)(2x^2-7y^9-4) ?
The answer to (\partial)/(\partial x)(2x^2-7y^9-4) is 4x