What is the tangent of f(x)=2^x,\at x=0 ?

The tangent of f(x)=2^x,\at x=0 is y=ln(2)x+1

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tangent of f(x)=2^x,\at x=0

tangent of

f (x) = 2^{x}, at x = 0

y = ln (2) x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

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 = ln (2)

Find the line with slope m=

ln (2)

and passing through

(0, 1) : y = ln (2) x + 1

y = ln (2) x + 1

\int \frac{1}{9 + x ^{2}} d x

y = \frac{- 5}{3 x}

f (x) = 2 x^{2} - 4 x - 5, at x = 2

\int sin (e^{x}) e^{x} d x

\frac{d y}{d x} = \frac{3 y ^{2} + 10 x ^{2}}{x y}

What is the tangent of f(x)=2^x,\at x=0 ?
The tangent of f(x)=2^x,\at x=0 is y=ln(2)x+1