What is (\partial)/(\partial y)(-1/y) ?

The answer to (\partial)/(\partial y)(-1/y) is 1/(y^2)

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(\partial)/(\partial y)(-1/y)

\frac{\partial}{\partial y} (- \frac{1}{y})

\frac{1}{y ^{2}}

Solution steps

\frac{\partial}{\partial y} (- \frac{1}{y})

Take the constant out:

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

= - \frac{\partial}{\partial y} (\frac{1}{y})

Apply exponent rule:

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

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= - (- 1 \cdot y^{- 1 - 1})

Simplify

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

= \frac{1}{y ^{2}}

\frac{d y}{d t} = y (1 - 0.0002 y)

\int 2^{x} x d x

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (π) = 0, y^{^{'}} (π) = 2

f (x) = 3 x^{2}, at x = 4

\frac{d}{d x} (\frac{x + 5}{x})

What is (\partial)/(\partial y)(-1/y) ?
The answer to (\partial)/(\partial y)(-1/y) is 1/(y^2)