What is the normal of y=x^4+5e^x,(0,5) ?

The normal of y=x^4+5e^x,(0,5) is y=-1/5 x+5

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normal of y=x^4+5e^x,(0,5)

normal of

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

y = - \frac{1}{5} x + 5

Solution steps

Find the slope of

y = x^{4} + 5 e^{x} : \frac{d y}{d x} = 4 x^{3} + 5 e^{x}

Compute the slope of

y = x^{4} + 5 e^{x}

(0, 5) : m = 5

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5}

Find the line with slope m=

- \frac{1}{5}

and passing through

(0, 5) : y = - \frac{1}{5} x + 5

y = - \frac{1}{5} x + 5

\frac{d y}{d x} = π y + 12 e

x \to π lim (\frac{5 x}{3})

y = x, y = x

3^{x^{2} - 2 x + 1}

\int y^{2} e^{- y} d y

What is the normal of y=x^4+5e^x,(0,5) ?
The normal of y=x^4+5e^x,(0,5) is y=-1/5 x+5