What is (\partial)/(\partial y)(2x(5+y)^{-1}) ?

The answer to (\partial)/(\partial y)(2x(5+y)^{-1}) is -(2x)/((5+y)^2)

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(\partial)/(\partial y)(2x(5+y)^{-1})

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

- \frac{2 x}{( 5 + y ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

Treat

x

as a constant

Take the constant out:

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

= 2 x \frac{\partial}{\partial y} ((5 + y)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 + y ) ^{2}} \frac{\partial}{\partial y} (5 + y)

= - \frac{1}{( 5 + y ) ^{2}} \frac{\partial}{\partial y} (5 + y)

\frac{\partial}{\partial y} (5 + y) = 1

= 2 x (- \frac{1}{( 5 + y ) ^{2}} \cdot 1)

Simplify

2 x (- \frac{1}{( 5 + y ) ^{2}} \cdot 1) : - \frac{2 x}{( 5 + y ) ^{2}}

= - \frac{2 x}{( 5 + y ) ^{2}}

\frac{sec ( 9 x ^{2} )}{tan ( 2 x ) \cdot cos ( x )}

y^{^{'}} = 3

1 + 6 x^{\frac{3}{2}}

\frac{d}{d x} (\frac{6 e ^{x}}{5 e ^{x} + 7})

\frac{d}{d x} (a^{x} + x^{a})

What is (\partial)/(\partial y)(2x(5+y)^{-1}) ?
The answer to (\partial)/(\partial y)(2x(5+y)^{-1}) is -(2x)/((5+y)^2)