What is the normal of y=7xe^x,(0,0) ?

The normal of y=7xe^x,(0,0) is y=-1/7 x

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normal of y=7xe^x,(0,0)

normal of

y = 7 x e^{x}, (0, 0)

y = - \frac{1}{7} x

Solution steps

Find the slope of

y = 7 x e^{x} : \frac{d y}{d x} = 7 (e^{x} + e^{x} x)

Compute the slope of

y = 7 x e^{x}

(0, 0) : m = 7

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(0, 0) : y = - \frac{1}{7} x

y = - \frac{1}{7} x

y = \frac{1}{t ( 8 t + 1 )}

2 x^{2}, 5 x - 3 x^{2}

y^{^{'}} = (1 - 7 x) y^{2}

\int tan^{5} (2 x) d x