What is (\partial)/(\partial x)(x^4+y^4-32x+4y+4) ?

The answer to (\partial)/(\partial x)(x^4+y^4-32x+4y+4) is 4x^3-32

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(\partial)/(\partial x)(x^4+y^4-32x+4y+4)

\frac{\partial}{\partial x} (x^{4} + y^{4} - 32 x + 4 y + 4)

4 x^{3} - 32

Solution steps

\frac{\partial}{\partial x} (x^{4} + y^{4} - 32 x + 4 y + 4)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4}) + \frac{\partial}{\partial x} (y^{4}) - \frac{\partial}{\partial x} (32 x) + \frac{\partial}{\partial x} (4 y) + \frac{\partial}{\partial x} (4)

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

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

\frac{\partial}{\partial x} (32 x) = 32

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

\frac{\partial}{\partial x} (4) = 0

= 4 x^{3} + 0 - 32 + 0 + 0

Simplify

= 4 x^{3} - 32

\frac{d}{d x} (f (x) (x g (x)))

y^{^{'}} + \frac{1}{t} y = 3 cos (2 t)

\frac{d}{d x} (x^{2} cos (ln (x)))

x \to 0 + lim (x \cdot e^{\frac{1}{x}})

\int \frac{x}{1 - 3 x ^{2}} d x

What is (\partial)/(\partial x)(x^4+y^4-32x+4y+4) ?
The answer to (\partial)/(\partial x)(x^4+y^4-32x+4y+4) is 4x^3-32