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

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

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

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

6 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 6 y + 0 - 0

Simplify

= 6 y

\frac{d ^{3}}{d x ^{3}} (e^{x})

\int cos^{4} (5 x + 5) d x

\int lo g_{2} (x - 4) d x

\frac{4 x ^{2} + 8 x + 4}{x}

x \to 0 lim (\frac{( x ^{3} )}{x ^{2}})

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