What is (\partial)/(\partial x)(x^2+2xy+3y^4) ?

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

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(\partial)/(\partial x)(x^2+2xy+3y^4)

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{4})

2 x + 2 y

Solution steps

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{4})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (2 x y) + \frac{\partial}{\partial x} (3 y^{4})

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

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

\frac{\partial}{\partial x} (3 y^{4}) = 0

= 2 x + 2 y + 0

Simplify

= 2 x + 2 y

\int \frac{1}{x x + 1} d x

y = ln (\frac{4}{x ^{3}})

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

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

\int \frac{54}{1 - cos ( 6 x )} d x

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