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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} - y = 2 t e^{2 t}, y (0) = 1

sin (x y)

\int (\frac{5}{x} - 2 x^{3}) d x

\frac{d y}{d x} - e^{5 x + y} = 0

\frac{d}{d x} (\frac{- 15}{x})

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