What is the derivative of (2x/(-8x-1)) ?

The derivative of (2x/(-8x-1)) is -2/((-8x-1)^2)

What is the first derivative of (2x/(-8x-1)) ?

The first derivative of (2x/(-8x-1)) is -2/((-8x-1)^2)

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\frac{d}{d x} (\frac{2 x}{- 8 x - 1})

- \frac{2}{( - 8 x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x}{- 8 x - 1})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{- 8 x - 1})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d x}{d x} ( - 8 x - 1 ) - \frac{d}{d x} ( - 8 x - 1 ) x}{( - 8 x - 1 ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (- 8 x - 1) = - 8

= 2 \cdot \frac{1 \cdot ( - 8 x - 1 ) - ( - 8 ) x}{( - 8 x - 1 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( - 8 x - 1 ) - ( - 8 ) x}{( - 8 x - 1 ) ^{2}} : - \frac{2}{( - 8 x - 1 ) ^{2}}

= - \frac{2}{( - 8 x - 1 ) ^{2}}

y^{^{'}} - \frac{2}{t} y = t^{2} e^{t}

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

\frac{d}{d y} (x^{6 y})

x \to - 3 lim (\frac{x + 3}{x - 3})

\frac{d}{d x} ((2 x + 1)^{4})

What is the derivative of (2x/(-8x-1)) ?
The derivative of (2x/(-8x-1)) is -2/((-8x-1)^2)
What is the first derivative of (2x/(-8x-1)) ?
The first derivative of (2x/(-8x-1)) is -2/((-8x-1)^2)