What is the normal of y=2xe^x,(0,0) ?

The normal of y=2xe^x,(0,0) is y=-1/2 x

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normal of y=2xe^x,(0,0)

normal of

y = 2 x e^{x}, (0, 0)

y = - \frac{1}{2} x

Solution steps

Find the slope of

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

Compute the slope of

y = 2 x e^{x}

(0, 0) : m = 2

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(0, 0) : y = - \frac{1}{2} x

y = - \frac{1}{2} x

cos (π x)

\frac{d y}{d x} + y cos (x) = 3 cos (x), y (0) = 5

y = 4 x - x^{2}, y = x

f (x) = \frac{5}{x}, at x = \frac{1}{16}

\int \frac{( 2 t ^{2} + t t - 1 )}{t ^{2}} d t