What is the normal of y=e^{2x},\at x=ln(2) ?

The normal of y=e^{2x},\at x=ln(2) is y=-1/8 x+4+1/8 ln(2)

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normal of y=e^{2x},\at x=ln(2)

normal of

y = e^{2 x}, at x = ln (2)

y = - \frac{1}{8} x + 4 + \frac{1}{8} ln (2)

Solution steps

Find the tangent point:

(ln (2), 4)

Find the slope of

y = e^{2 x} : \frac{d y}{d x} = e^{2 x} \cdot 2

Compute the slope of

y = e^{2 x}

x = ln (2) : m = 8

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{8}

Find the line with slope m=

- \frac{1}{8}

and passing through

(ln (2), 4) : y = - \frac{1}{8} x + 4 + \frac{1}{8} ln (2)

y = - \frac{1}{8} x + 4 + \frac{1}{8} ln (2)

\frac{\partial}{\partial y} (y cos (x y) + e^{x})

\int cos^{2} (t) - sin^{2} (t) d t

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))

(- 2 cos (2 x) - sin (3 x))^{2}

What is the normal of y=e^{2x},\at x=ln(2) ?
The normal of y=e^{2x},\at x=ln(2) is y=-1/8 x+4+1/8 ln(2)