What is the derivative of (1-6x/(5+x)) ?

The derivative of (1-6x/(5+x)) is -(31)/((5+x)^2)

What is the first derivative of (1-6x/(5+x)) ?

The first derivative of (1-6x/(5+x)) is -(31)/((5+x)^2)

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\frac{d}{d x} (\frac{1 - 6 x}{5 + x})

- \frac{31}{( 5 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 - 6 x}{5 + x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1 - 6 x ) ( 5 + x ) - \frac{d}{d x} ( 5 + x ) ( 1 - 6 x )}{( 5 + x ) ^{2}}

\frac{d}{d x} (1 - 6 x) = - 6

\frac{d}{d x} (5 + x) = 1

= \frac{( - 6 ) ( 5 + x ) - 1 \cdot ( 1 - 6 x )}{( 5 + x ) ^{2}}

Simplify

\frac{( - 6 ) ( 5 + x ) - 1 \cdot ( 1 - 6 x )}{( 5 + x ) ^{2}} : - \frac{31}{( 5 + x ) ^{2}}

= - \frac{31}{( 5 + x ) ^{2}}

\int \frac{1}{8 - x} d x

\frac{d}{d x} (7 x + 8)

z e^{z}

n = 0 \sum \infty (\frac{n}{n + 3})^{n}

What is the derivative of (1-6x/(5+x)) ?
The derivative of (1-6x/(5+x)) is -(31)/((5+x)^2)
What is the first derivative of (1-6x/(5+x)) ?
The first derivative of (1-6x/(5+x)) is -(31)/((5+x)^2)