What is the derivative of ln((x^2+4/(x^2+1))) ?

The derivative of ln((x^2+4/(x^2+1))) is -(6x)/((x^2+4)(x^2+1))

What is the first derivative of ln((x^2+4/(x^2+1))) ?

The first derivative of ln((x^2+4/(x^2+1))) is -(6x)/((x^2+4)(x^2+1))

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

- \frac{6 x}{( x ^{2} + 4 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} + 4}{x ^{2} + 1}) = - \frac{6 x}{( x ^{2} + 1 ) ^{2}}

= \frac{1}{\frac{x ^{2} + 4}{x ^{2} + 1}} (- \frac{6 x}{( x ^{2} + 1 ) ^{2}})

Simplify

\frac{1}{\frac{x ^{2} + 4}{x ^{2} + 1}} (- \frac{6 x}{( x ^{2} + 1 ) ^{2}}) : - \frac{6 x}{( x ^{2} + 4 ) ( x ^{2} + 1 )}

= - \frac{6 x}{( x ^{2} + 4 ) ( x ^{2} + 1 )}

x^{3} - 6 x^{2} - 15 x + 4

x \to 5 lim (x^{- 2})

y^{^{'}} = \frac{9 x}{y + x ^{2} y}

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

\frac{y ^{^{'}}}{2} = y

What is the derivative of ln((x^2+4/(x^2+1))) ?
The derivative of ln((x^2+4/(x^2+1))) is -(6x)/((x^2+4)(x^2+1))
What is the first derivative of ln((x^2+4/(x^2+1))) ?
The first derivative of ln((x^2+4/(x^2+1))) is -(6x)/((x^2+4)(x^2+1))