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

The answer to (\partial)/(\partial x)(x^3+4x^2y+4x-3y+2) is 3x^2+8xy+4

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

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

3 x^{2} + 8 x y + 4

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

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

= 3 x^{2} + 8 y x + 4 - 0 + 0

Simplify

= 3 x^{2} + 8 x y + 4

x \to 0 lim (\frac{x ^{2} - 8 x + 15}{5})

y^{2} y^{^{'}} + y^{3} = 1, y (0) = 0

\int_{2}^{\infty} e^{- 2 x} d x

f (x) = x cos (x^{2})

\frac{\partial}{\partial y} (x^{3} - 6 x y + y^{3})

What is (\partial)/(\partial x)(x^3+4x^2y+4x-3y+2) ?
The answer to (\partial)/(\partial x)(x^3+4x^2y+4x-3y+2) is 3x^2+8xy+4