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

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

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

tangent of

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

y = 2 x + 2 - 2 ln (2)

Solution steps

Find the tangent point:

(ln (2), 2)

Find the slope of

f (x) = e^{x} : \frac{df ( x )}{d x} = 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 = 2

Find the line with slope m=

2

and passing through

(ln (2), 2) : y = 2 x + 2 - 2 ln (2)

y = 2 x + 2 - 2 ln (2)

\frac{\partial}{\partial x} (2 x - y e^{y})

in (2 x), (0, 2)

\frac{d y}{d x} = 3 (1 + y) (1 - x)

\int \frac{1}{x 36 x ^{2} - 36} d x

\int \frac{x}{x + 15} d x

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