What is (\partial)/(\partial x)((2y)/(9x^2+y^2+9)) ?

The answer to (\partial)/(\partial x)((2y)/(9x^2+y^2+9)) is -(36yx)/((9x^2+y^2+9)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((2y)/(9x^2+y^2+9))

\frac{\partial}{\partial x} (\frac{2 y}{9 x ^{2} + y ^{2} + 9})

- \frac{36 y x}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{2 y}{9 x ^{2} + y ^{2} + 9})

Treat

y

as a constant

Take the constant out:

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

= 2 y \frac{\partial}{\partial x} (\frac{1}{9 x ^{2} + y ^{2} + 9})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 2 y \frac{\partial}{\partial x} ((9 x^{2} + y^{2} + 9)^{- 1})

Apply the chain rule:

- \frac{1}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}} \frac{\partial}{\partial x} (9 x^{2} + y^{2} + 9)

= - \frac{1}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}} \frac{\partial}{\partial x} (9 x^{2} + y^{2} + 9)

\frac{\partial}{\partial x} (9 x^{2} + y^{2} + 9) = 18 x

= 2 y (- \frac{1}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}} \cdot 18 x)

Simplify

2 y (- \frac{1}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}} \cdot 18 x) : - \frac{36 y x}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}}

= - \frac{36 y x}{( 9 x ^{2} + y ^{2} + 9 ) ^{2}}

ln (x^{2} + 9)

\frac{d}{d x} (2 e^{- x^{2}})

\int 6 cos (3 x) d x

\int \frac{1}{x ( 1 - 0.0004 x )} d x

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

What is (\partial)/(\partial x)((2y)/(9x^2+y^2+9)) ?
The answer to (\partial)/(\partial x)((2y)/(9x^2+y^2+9)) is -(36yx)/((9x^2+y^2+9)^2)