What is the derivative of f(x)=((1-4x))/(1+4x) ?

The derivative of f(x)=((1-4x))/(1+4x) is -8/((1+4x)^2)

What is the first derivative of f(x)=((1-4x))/(1+4x) ?

The first derivative of f(x)=((1-4x))/(1+4x) is -8/((1+4x)^2)

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derivative of

f (x) = \frac{( 1 - 4 x )}{1 + 4 x}

- \frac{8}{( 1 + 4 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 - 4 x}{1 + 4 x})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (1 - 4 x) = - 4

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

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

Simplify

\frac{( - 4 ) ( 1 + 4 x ) - 4 ( 1 - 4 x )}{( 1 + 4 x ) ^{2}} : - \frac{8}{( 1 + 4 x ) ^{2}}

= - \frac{8}{( 1 + 4 x ) ^{2}}

t \frac{d y}{d t} = 4 y

n = 0 \sum \infty (\frac{e}{8 π})^{n}

\int \frac{ln ( 2 x )}{x ln ( x )} d x

x \to 1 lim (\frac{1 - x ^{2}}{x - 1})

y = \frac{2}{x - 8} (10.1)

What is the derivative of f(x)=((1-4x))/(1+4x) ?
The derivative of f(x)=((1-4x))/(1+4x) is -8/((1+4x)^2)
What is the first derivative of f(x)=((1-4x))/(1+4x) ?
The first derivative of f(x)=((1-4x))/(1+4x) is -8/((1+4x)^2)