What is the derivative of (100x/(x^2+x+4)) ?

The derivative of (100x/(x^2+x+4)) is (100(-x^2+4))/((x^2+x+4)^2)

What is the first derivative of (100x/(x^2+x+4)) ?

The first derivative of (100x/(x^2+x+4)) is (100(-x^2+4))/((x^2+x+4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{100 x}{x ^{2} + x + 4})

\frac{100 ( - x ^{2} + 4 )}{( x ^{2} + x + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{100 x}{x ^{2} + x + 4})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 100 \frac{d}{d x} (\frac{x}{x ^{2} + x + 4})

Apply the Quotient Rule:

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

= 100 \cdot \frac{\frac{d x}{d x} ( x ^{2} + x + 4 ) - \frac{d}{d x} ( x ^{2} + x + 4 ) x}{( x ^{2} + x + 4 ) ^{2}}

\frac{d x}{d x} = 1

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

= 100 \cdot \frac{1 \cdot ( x ^{2} + x + 4 ) - ( 2 x + 1 ) x}{( x ^{2} + x + 4 ) ^{2}}

Simplify

100 \cdot \frac{1 \cdot ( x ^{2} + x + 4 ) - ( 2 x + 1 ) x}{( x ^{2} + x + 4 ) ^{2}} : \frac{100 ( - x ^{2} + 4 )}{( x ^{2} + x + 4 ) ^{2}}

= \frac{100 ( - x ^{2} + 4 )}{( x ^{2} + x + 4 ) ^{2}}

f (x) = \frac{5}{x}

\int x^{- \frac{3}{4}} d x

\int_{0}^{π} sin^{2} (3 x) d x

y^{^{''}} + 121 y = sec (11 x), y (0) = - 5, y^{^{'}} (0) = 6

\frac{\partial}{\partial y} (3 x (y - 1)^{\frac{1}{3}})

What is the derivative of (100x/(x^2+x+4)) ?
The derivative of (100x/(x^2+x+4)) is (100(-x^2+4))/((x^2+x+4)^2)
What is the first derivative of (100x/(x^2+x+4)) ?
The first derivative of (100x/(x^2+x+4)) is (100(-x^2+4))/((x^2+x+4)^2)