What is the tangent of f(x)=ln((x+2)/(x^2)),\at x=1 ?

The tangent of f(x)=ln((x+2)/(x^2)),\at x=1 is y=-5/3 x+ln(3)+5/3

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tangent of f(x)=ln((x+2)/(x^2)),\at x=1

tangent of

f (x) = ln (\frac{x + 2}{x ^{2}}), at x = 1

y = - \frac{5}{3} x + ln (3) + \frac{5}{3}

Solution steps

Find the tangent point:

(1, ln (3))

Find the slope of

f (x) = ln (\frac{x + 2}{x ^{2}}) : \frac{df ( x )}{d x} = \frac{- x - 4}{x ( x + 2 )}

Compute the slope of

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

x = 1 : m = - \frac{5}{3}

Find the line with slope m=

- \frac{5}{3}

and passing through

(1, ln (3)) : y = - \frac{5}{3} x + ln (3) + \frac{5}{3}

y = - \frac{5}{3} x + ln (3) + \frac{5}{3}

\frac{d}{d x} (\frac{( ln ( x ) ) ^{3}}{x ^{3}})

x \to 0 lim (x^{2} - 3 x + 1)

\int \frac{ln ∣ x ∣}{x} d x

n = 1 \sum \infty \frac{1}{( 4 n ) ^{2} - 1}

x \to - 1 + lim (- 1)

What is the tangent of f(x)=ln((x+2)/(x^2)),\at x=1 ?
The tangent of f(x)=ln((x+2)/(x^2)),\at x=1 is y=-5/3 x+ln(3)+5/3