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

The answer to (\partial)/(\partial x)(3x^2-3y^2) is 6x

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

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

6 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

= 6 x - 0

Simplify

= 6 x

x \to - 4 lim (- 5 x)

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

x^{2} y^{^{''}} - 5 x y^{^{'}} + 8 y = 0

f (x) = 2 sin (x) + 1, at x = - \frac{π}{2}

\int 5 x e^{x^{2}} d x

What is (\partial)/(\partial x)(3x^2-3y^2) ?
The answer to (\partial)/(\partial x)(3x^2-3y^2) is 6x