What is (\partial)/(\partial x)((x^2+y^2)^{1/3}) ?

The answer to (\partial)/(\partial x)((x^2+y^2)^{1/3}) is (2x)/(3(x^2+y^2)^{2/3)}

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(\partial)/(\partial x)((x^2+y^2)^{1/3})

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{3}})

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

Solution steps

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{3}})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}} \frac{\partial}{\partial x} (x^{2} + y^{2})

= \frac{1}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}} \frac{\partial}{\partial x} (x^{2} + y^{2})

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

= \frac{1}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}} \cdot 2 x

Simplify

\frac{1}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}} \cdot 2 x : \frac{2 x}{3 ( x ^{2} + y ^{2} ) ^{\frac{2}{3}}}

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

f (x) = arcsin (5 x)

x \to 1 lim (\frac{- 1}{x - 1})

2 y^{^{''}} + y^{^{'}} - 3 y = 0, y (0) = a, y^{^{'}} (0) = b

\frac{d}{d x} (\frac{1}{x ^{7} + 2})

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

What is (\partial)/(\partial x)((x^2+y^2)^{1/3}) ?
The answer to (\partial)/(\partial x)((x^2+y^2)^{1/3}) is (2x)/(3(x^2+y^2)^{2/3)}