What is the derivative of (-2x-y)/(x+3y^2) ?

The derivative of (-2x-y)/(x+3y^2) is (-6y^2+y)/((x+3y^2)^2)

What is the first derivative of (-2x-y)/(x+3y^2) ?

The first derivative of (-2x-y)/(x+3y^2) is (-6y^2+y)/((x+3y^2)^2)

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derivative of

\frac{- 2 x - y}{x + 3 y ^{2}}

\frac{- 6 y ^{2} + y}{( x + 3 y ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{- 2 x - y}{x + 3 y ^{2}})

Treat

y

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( - 2 x - y ) ( x + 3 y ^{2} ) - \frac{d}{d x} ( x + 3 y ^{2} ) ( - 2 x - y )}{( x + 3 y ^{2} ) ^{2}}

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

\frac{d}{d x} (x + 3 y^{2}) = 1

= \frac{( - 2 ) ( x + 3 y ^{2} ) - 1 \cdot ( - 2 x - y )}{( x + 3 y ^{2} ) ^{2}}

Simplify

\frac{( - 2 ) ( x + 3 y ^{2} ) - 1 \cdot ( - 2 x - y )}{( x + 3 y ^{2} ) ^{2}} : \frac{- 6 y ^{2} + y}{( x + 3 y ^{2} ) ^{2}}

= \frac{- 6 y ^{2} + y}{( x + 3 y ^{2} ) ^{2}}

\frac{d y}{d x} = \frac{( x - 3 )}{( y - 3 )}

\int (x + 4) (x - 2) d x

y = 3 sin (x), at x = \frac{π}{6}

\frac{d y}{d x}, e^{x} y = 2

\int (2 x + 5 - \frac{1}{x}) d x

What is the derivative of (-2x-y)/(x+3y^2) ?
The derivative of (-2x-y)/(x+3y^2) is (-6y^2+y)/((x+3y^2)^2)
What is the first derivative of (-2x-y)/(x+3y^2) ?
The first derivative of (-2x-y)/(x+3y^2) is (-6y^2+y)/((x+3y^2)^2)