What is the derivative of y=(-3-5y)/(-8y+5x) ?

The derivative of y=(-3-5y)/(-8y+5x) is (-25x-24)/((-8y+5x)^2)

What is the first derivative of y=(-3-5y)/(-8y+5x) ?

The first derivative of y=(-3-5y)/(-8y+5x) is (-25x-24)/((-8y+5x)^2)

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derivative of

y = \frac{- 3 - 5 y}{- 8 y + 5 x}

\frac{- 25 x - 24}{( - 8 y + 5 x ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{- 3 - 5 y}{- 8 y + 5 x})

Treat

x

as a constant

Apply the Quotient Rule:

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

= \frac{\frac{d}{d y} ( - 3 - 5 y ) ( - 8 y + 5 x ) - \frac{d}{d y} ( - 8 y + 5 x ) ( - 3 - 5 y )}{( - 8 y + 5 x ) ^{2}}

\frac{d}{d y} (- 3 - 5 y) = - 5

\frac{d}{d y} (- 8 y + 5 x) = - 8

= \frac{( - 5 ) ( - 8 y + 5 x ) - ( - 8 ) ( - 3 - 5 y )}{( - 8 y + 5 x ) ^{2}}

Simplify

\frac{( - 5 ) ( - 8 y + 5 x ) - ( - 8 ) ( - 3 - 5 y )}{( - 8 y + 5 x ) ^{2}} : \frac{- 25 x - 24}{( - 8 y + 5 x ) ^{2}}

= \frac{- 25 x - 24}{( - 8 y + 5 x ) ^{2}}

(t^{2} - x t^{2}) \frac{d x}{d t} + x^{2} + t x^{2} = 0

\int_{0}^{1} x e^{- 3 x} d x

\int \frac{5 t}{1 - 25 t ^{4}} d t

cot (csc (x))

g (x) = \frac{1}{x + 4}

What is the derivative of y=(-3-5y)/(-8y+5x) ?
The derivative of y=(-3-5y)/(-8y+5x) is (-25x-24)/((-8y+5x)^2)
What is the first derivative of y=(-3-5y)/(-8y+5x) ?
The first derivative of y=(-3-5y)/(-8y+5x) is (-25x-24)/((-8y+5x)^2)