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

The answer to (\partial)/(\partial x)(3x^3-y^2+4xy) is 9x^2+4y

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

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

9 x^{2} + 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{3}) = 9 x^{2}

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

\frac{\partial}{\partial x} (4 x y) = 4 y

= 9 x^{2} - 0 + 4 y

Simplify

= 9 x^{2} + 4 y

y^{^{''}} + y^{^{'}} + y = t + e^{2 t}

y^{^{'}} + 2 y = 1 + e^{3 x}

4 x, x, x^{3}

\frac{\partial}{\partial x} ((x^{2} + 3 y^{2})^{7})

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

What is (\partial)/(\partial x)(3x^3-y^2+4xy) ?
The answer to (\partial)/(\partial x)(3x^3-y^2+4xy) is 9x^2+4y