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

Compute the slope of

y = 2 + 1 x + 10 x e^{x}

x = 0 : m = 11

Find the line with slope m=

11

and passing through

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

y = 11 x + 2

y^{^{''}} + y = cos (2 x)

\int (2 - x)^{2} d x

3 y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = - 7, y^{^{'}} (0) = - 8

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

y = 4 x^{2} + 5 x + 1, (0, 1)

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