What is (\partial)/(\partial y)(5x(4+y)^{-1}) ?

The answer to (\partial)/(\partial y)(5x(4+y)^{-1}) is -(5x)/((4+y)^2)

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(\partial)/(\partial y)(5x(4+y)^{-1})

\frac{\partial}{\partial y} (5 x (4 + y)^{- 1})

- \frac{5 x}{( 4 + y ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (5 x (4 + y)^{- 1})

Treat

x

as a constant

Take the constant out:

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

= 5 x \frac{\partial}{\partial y} ((4 + y)^{- 1})

Apply the chain rule:

- \frac{1}{( 4 + y ) ^{2}} \frac{\partial}{\partial y} (4 + y)

= - \frac{1}{( 4 + y ) ^{2}} \frac{\partial}{\partial y} (4 + y)

\frac{\partial}{\partial y} (4 + y) = 1

= 5 x (- \frac{1}{( 4 + y ) ^{2}} \cdot 1)

Simplify

5 x (- \frac{1}{( 4 + y ) ^{2}} \cdot 1) : - \frac{5 x}{( 4 + y ) ^{2}}

= - \frac{5 x}{( 4 + y ) ^{2}}

5 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

f (x) = \frac{16}{x}

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

f (x) = sin^{2} (x^{2})

x \to π lim (cot (x))

What is (\partial)/(\partial y)(5x(4+y)^{-1}) ?
The answer to (\partial)/(\partial y)(5x(4+y)^{-1}) is -(5x)/((4+y)^2)