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}}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 2)

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

\frac{\partial}{\partial x} (6 e^{x^{5} y})

\int_{0}^{1} \frac{x - 1}{x} d x

g (x) = - 3 x^{2} + 5 x - 6

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)