What is the derivative of y=ln(x/(x^2+4)) ?

The derivative of y=ln(x/(x^2+4)) is (-x^2+4)/(x(x^2+4))

What is the first derivative of y=ln(x/(x^2+4)) ?

The first derivative of y=ln(x/(x^2+4)) is (-x^2+4)/(x(x^2+4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = ln (\frac{x}{x ^{2} + 4})

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = \frac{1}{x}, y = 0, x = 1, x = e

\int \frac{5}{( 81 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{4}{x ^{6}}

\frac{\partial}{\partial x} (\frac{x}{x ^{7} - y ^{5}})

x \to 0 + lim (e^{3 x l n (x)})

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