What is the d/(dy)((2x)/y) ?

The d/(dy)((2x)/y) is -(2x)/(y^2)

What is the first d/(dy)((2x)/y) ?

The first d/(dy)((2x)/y) is -(2x)/(y^2)

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\frac{d}{d y} (\frac{2 x}{y})

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

Solution steps

\frac{d}{d y} (\frac{2 x}{y})

Treat

x

as a constant

Take the constant out:

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

= 2 x \frac{d}{d y} (\frac{1}{y})

Apply exponent rule:

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

= 2 x \frac{d}{d y} (y^{- 1})

Apply the Power Rule:

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

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

Simplify

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

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

\frac{s - 3}{( s - 4 ) ( s + 6 )}

\int \frac{x + 1}{x ^{3} + 7 x ^{2} + 6 x} d x

\frac{d}{d x} (\frac{4}{5 x ^{6}})

f (x) = 2 - 6 x^{2}, (- 5, - 148)

\frac{d}{d x} ((3 x^{4} - 6 x)^{3})