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

The derivative of f(x)=(3-(1/(x^2)))/(x+4) is (-3x^3+3x+8)/(x^3(x+4)^2)

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

The first derivative of f(x)=(3-(1/(x^2)))/(x+4) is (-3x^3+3x+8)/(x^3(x+4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{- 3 x ^{3} + 3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{\frac{2}{x ^{3}} ( x + 4 ) - 1 \cdot ( 3 - \frac{1}{x ^{2}} )}{( x + 4 ) ^{2}} : \frac{- 3 x ^{3} + 3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

= \frac{- 3 x ^{3} + 3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

\int e^{- s t} cos (t) d t

t y^{^{'}} = 2

\frac{d}{d x} (sec^{2} (x))

\frac{d}{d x} (\frac{sin ( x )}{8 x e ^{x}})

\frac{d}{d t} (t^{5})

What is the derivative of f(x)=(3-(1/(x^2)))/(x+4) ?
The derivative of f(x)=(3-(1/(x^2)))/(x+4) is (-3x^3+3x+8)/(x^3(x+4)^2)
What is the first derivative of f(x)=(3-(1/(x^2)))/(x+4) ?
The first derivative of f(x)=(3-(1/(x^2)))/(x+4) is (-3x^3+3x+8)/(x^3(x+4)^2)