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

The answer to (\partial)/(\partial y)(xsqrt(x^2+y^2)) is (xy)/(sqrt(x^2+y^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(xsqrt(x^2+y^2))

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

\frac{x y}{x ^{2} + y ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{x y}{x ^{2} + y ^{2}}

\frac{d}{d x} (cos (x) + 2 sin (x))

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

\frac{\partial}{\partial y} (8 y - (x + y)^{3})

2 sin^{4} (x)

\int \frac{1}{( 2 - y ) ( 1 - y )} d y

What is (\partial)/(\partial y)(xsqrt(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(xsqrt(x^2+y^2)) is (xy)/(sqrt(x^2+y^2))