What is the tangent of f(x)=3^x,\at x=0 ?

The tangent of f(x)=3^x,\at x=0 is y=ln(3)x+1

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tangent of f(x)=3^x,\at x=0

tangent of

f (x) = 3^{x}, at x = 0

y = ln (3) x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = 3^{x} : \frac{df ( x )}{d x} = 3^{x} ln (3)

Compute the slope of

f (x) = 3^{x}

x = 0 : m = ln (3)

Find the line with slope m=

ln (3)

and passing through

(0, 1) : y = ln (3) x + 1

y = ln (3) x + 1

\int - 9 sin (3 t) d t

x \to 1 lim (x^{17} - x + 3)

\frac{d}{d x} (3 x^{2} + 2 y^{4})

\frac{d}{d y} (e^{- x y})

What is the tangent of f(x)=3^x,\at x=0 ?
The tangent of f(x)=3^x,\at x=0 is y=ln(3)x+1