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

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

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

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

2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (3) = 0

= 0 + 2 x - 0

Simplify

= 2 x

y^{^{'}} = cos (2 x)

25 cos (\frac{π x}{50})

\int cos^{3} (a x + b) d x

x \to \infty lim (e^{2})

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