What is the derivative of ln(x/(x^2+5)) ?

The derivative of ln(x/(x^2+5)) is (-x^2+5)/(x(x^2+5))

What is the first derivative of ln(x/(x^2+5)) ?

The first derivative of ln(x/(x^2+5)) is (-x^2+5)/(x(x^2+5))

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\frac{d}{d x} (ln (\frac{x}{x ^{2} + 5}))

\frac{- x ^{2} + 5}{x ( x ^{2} + 5 )}

Solution steps

\frac{d}{d x} (ln (\frac{x}{x ^{2} + 5}))

Apply the chain rule:

\frac{1}{\frac{x}{x ^{2} + 5}} \frac{d}{d x} (\frac{x}{x ^{2} + 5})

= \frac{1}{\frac{x}{x ^{2} + 5}} \frac{d}{d x} (\frac{x}{x ^{2} + 5})

\frac{d}{d x} (\frac{x}{x ^{2} + 5}) = \frac{- x ^{2} + 5}{( x ^{2} + 5 ) ^{2}}

= \frac{1}{\frac{x}{x ^{2} + 5}} \cdot \frac{- x ^{2} + 5}{( x ^{2} + 5 ) ^{2}}

Simplify

\frac{1}{\frac{x}{x ^{2} + 5}} \cdot \frac{- x ^{2} + 5}{( x ^{2} + 5 ) ^{2}} : \frac{- x ^{2} + 5}{x ( x ^{2} + 5 )}

= \frac{- x ^{2} + 5}{x ( x ^{2} + 5 )}

\int - \frac{2}{t + 1} d t

9 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 2

f (x) = 5 - 2 x^{2}, at x = - 2

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

\frac{d}{d x} (sin (3 x + 2))

What is the derivative of ln(x/(x^2+5)) ?
The derivative of ln(x/(x^2+5)) is (-x^2+5)/(x(x^2+5))
What is the first derivative of ln(x/(x^2+5)) ?
The first derivative of ln(x/(x^2+5)) is (-x^2+5)/(x(x^2+5))