What is the tangent of y=e^x,\at x=0 ?

The tangent of y=e^x,\at x=0 is y=x+1

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tangent of y=e^x,\at x=0

tangent of

y = e^{x}, at x = 0

y = x + 1

Solution steps

Find the tangent point:

(0, 1)

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 = 1

Find the line with slope m=

1

and passing through

(0, 1) : y = x + 1

y = x + 1

\int \frac{12 x}{( x - 1 ) ( x ^{2} + 2 x + 3 )} d x

s l o p e in t erce pt (1.2) (3.3)

\frac{d P}{d t} = 4 Pt

\frac{d v}{d t} = k 3 v^{2}, v (0) = 27

\frac{d}{d x} (e^{s i n (x)} \cdot cos (x))