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

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

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

tangent of

y = 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

y = e^{x} : \frac{d y}{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)

x \to 1 lim (5 - x)

f (x) = 3 cos^{2} (x)

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

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

\frac{d y}{d x} = \frac{- y - 6 x ^{2}}{x + 3}

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