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

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

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

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

2 x

Solution steps

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

Treat

y, w

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 x + 0 + 0

Simplify

= 2 x

y = ∣ 3 x ∣, y = x^{2} - 4

x \to \infty lim (34)

f (x) = \frac{2 x}{x ^{2} + 1}

\int \frac{1}{1 - 4 x} d x

x \to - 1 lim (\frac{x ^{2} + 5}{x + 6})

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