What is (\partial)/(\partial y)(2x^2+4y^2+1/2) ?

The answer to (\partial)/(\partial y)(2x^2+4y^2+1/2) is 8y

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(\partial)/(\partial y)(2x^2+4y^2+1/2)

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

8 y

Solution steps

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (2 x^{2}) + \frac{\partial}{\partial y} (4 y^{2}) + \frac{\partial}{\partial y} (\frac{1}{2})

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

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

\frac{\partial}{\partial y} (\frac{1}{2}) = 0

= 0 + 8 y + 0

Simplify

= 8 y

y^{^{'}} = 1 + y

\int_{0}^{1} \frac{1}{x - 1} d x

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

\frac{4}{( 3 s - 1 ) ^{5}}

\int \frac{1}{5 + 4 sec ( x )} d x

What is (\partial)/(\partial y)(2x^2+4y^2+1/2) ?
The answer to (\partial)/(\partial y)(2x^2+4y^2+1/2) is 8y