What is the tangent of y=2+1x+10xe^x,\at x=0 ?

The tangent of y=2+1x+10xe^x,\at x=0 is y=11x+2

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tangent of y=2+1x+10xe^x,\at x=0

tangent of

y = 2 + 1 x + 10 x e^{x}, at x = 0

y = 11 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

y = 2 + 1 x + 10 x e^{x} : \frac{d y}{d x} = 10 (e^{x} x + e^{x}) + 1

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 = 11

Find the line with slope m=

11

and passing through

(0, 2) : y = 11 x + 2

y = 11 x + 2

What is the tangent of y=2+1x+10xe^x,\at x=0 ?
The tangent of y=2+1x+10xe^x,\at x=0 is y=11x+2