What is the derivative of f(x)=((1-4x))/(3+x) ?

The derivative of f(x)=((1-4x))/(3+x) is -(13)/((3+x)^2)

What is the first derivative of f(x)=((1-4x))/(3+x) ?

The first derivative of f(x)=((1-4x))/(3+x) is -(13)/((3+x)^2)

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derivative of

f (x) = \frac{( 1 - 4 x )}{3 + x}

- \frac{13}{( 3 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 - 4 x}{3 + 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 ) ( 3 + x ) - \frac{d}{d x} ( 3 + x ) ( 1 - 4 x )}{( 3 + x ) ^{2}}

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

\frac{d}{d x} (3 + x) = 1

= \frac{( - 4 ) ( 3 + x ) - 1 \cdot ( 1 - 4 x )}{( 3 + x ) ^{2}}

Simplify

\frac{( - 4 ) ( 3 + x ) - 1 \cdot ( 1 - 4 x )}{( 3 + x ) ^{2}} : - \frac{13}{( 3 + x ) ^{2}}

= - \frac{13}{( 3 + x ) ^{2}}

\frac{d}{d x} (1 0^{9 - x^{2}})

t^{2} sin (2 t)

x y^{^{'}} = y + 4 x^{2} sin (x), y (π) = 0

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

\frac{d}{d x} (x^{4} - 13 x^{2} + 36)

What is the derivative of f(x)=((1-4x))/(3+x) ?
The derivative of f(x)=((1-4x))/(3+x) is -(13)/((3+x)^2)
What is the first derivative of f(x)=((1-4x))/(3+x) ?
The first derivative of f(x)=((1-4x))/(3+x) is -(13)/((3+x)^2)