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

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

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

normal of

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

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

Solution steps

Find the tangent point:

(ln (2), 2)

Find the slope of

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

Compute the slope of

y = e^{x}

x = ln (2) : 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

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

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

\int x (2 x + 5)^{20} d x

\frac{\partial}{\partial x} (7 e^{x} sin (y))

x \to 2 lim (7 x - 4)

x \to 2 lim (\frac{x ^{3} + 1}{x + 1})

x \to 9 lim (\frac{e ^{5 x} - 1}{x})

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