What is (\partial)/(\partial x)(2x^2+3y^2-4x-5) ?

The answer to (\partial)/(\partial x)(2x^2+3y^2-4x-5) is 4x-4

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(\partial)/(\partial x)(2x^2+3y^2-4x-5)

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - 4 x - 5)

4 x - 4

Solution steps

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - 4 x - 5)

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} (3 y^{2}) - \frac{\partial}{\partial x} (4 x) - \frac{\partial}{\partial x} (5)

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

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

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

\frac{\partial}{\partial x} (5) = 0

= 4 x + 0 - 4 - 0

Simplify

= 4 x - 4

What is (\partial)/(\partial x)(2x^2+3y^2-4x-5) ?
The answer to (\partial)/(\partial x)(2x^2+3y^2-4x-5) is 4x-4