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

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

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

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

4 x - 2

Solution steps

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

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} (2 x)

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

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

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

= 4 x - 0 - 2

Simplify

= 4 x - 2

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

\int tan (\frac{x}{2}) d x

\frac{d}{d t} (t cos (6 t))

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

\int \frac{x ^{2} - 8 x - 1}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

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