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

The answer to (\partial)/(\partial x)(4x^3+4y^2x+4x) is 12x^2+4y^2+4

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

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

12 x^{2} + 4 y^{2} + 4

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (4 x^{3}) = 12 x^{2}

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

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

= 12 x^{2} + 4 y^{2} + 4

\int \frac{t}{t ^{4} + 2} d t

\frac{- 4 s + 8}{s ^{2} + 6 s + 8}

x \to \infty lim (7 - 2 x)

\frac{\partial}{\partial t} (e^{- 4 t} cos (π x))

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 6 x}{x + 1})

What is (\partial)/(\partial x)(4x^3+4y^2x+4x) ?
The answer to (\partial)/(\partial x)(4x^3+4y^2x+4x) is 12x^2+4y^2+4