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

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

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

tangent of

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

y = \frac{2}{3} x + ln (3) - \frac{2}{3} ln (2)

Solution steps

Find the tangent point:

(ln (2), ln (3))

Find the slope of

f (x) = ln (1 + e^{x}) : \frac{df ( x )}{d x} = \frac{e ^{x}}{1 + e ^{x}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

(ln (2), ln (3)) : y = \frac{2}{3} x + ln (3) - \frac{2}{3} ln (2)

y = \frac{2}{3} x + ln (3) - \frac{2}{3} ln (2)

\int x (x - 3)^{2} d x

y = x^{4} - 4 x^{2} + 4, y = x^{2}

g (t) = 2 t^{\frac{- 3}{4}}

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

6 ln (9 - x)

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