What is the tangent of y=e^{x/3},\at x=0 ?

The tangent of y=e^{x/3},\at x=0 is y= 1/3 x+1

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tangent of y=e^{x/3},\at x=0

tangent of

y = e^{\frac{x}{3}}, at x = 0

y = \frac{1}{3} x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

y = e^{\frac{x}{3}} : \frac{d y}{d x} = e^{\frac{x}{3}} \frac{1}{3}

Compute the slope of

y = e^{\frac{x}{3}}

x = 0 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(0, 1) : y = \frac{1}{3} x + 1

y = \frac{1}{3} x + 1

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

n = 0 \sum \infty 2 + e^{- 3 n}

g (s) = \frac{5}{s - 3}

\frac{d}{d x} (- 2 x cos (x))

\frac{5}{2} y^{4} y^{^{'}} = \frac{y ^{5} + 4}{x ^{3}}

What is the tangent of y=e^{x/3},\at x=0 ?
The tangent of y=e^{x/3},\at x=0 is y= 1/3 x+1