What is the normal of sqrt(1-tanh(5x)),\at x=0 ?

The normal of sqrt(1-tanh(5x)),\at x=0 is y= 2/5 x+1

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normal of sqrt(1-tanh(5x)),\at x=0

normal of

1 - tanh (5 x), at x = 0

y = \frac{2}{5} x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

y = 1 - tanh (5 x) : \frac{d y}{d x} = - \frac{5 sech ^{2} ( 5 x )}{2 1 - tanh ( 5 x )}

Compute the slope of

y = 1 - tanh (5 x)

x = 0 : m = - \frac{5}{2}

Compute the slope of the perpendicular line:

m_{p} = \frac{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(0, 1) : y = \frac{2}{5} x + 1

y = \frac{2}{5} x + 1

(3, - 3)

f (x) = 5 x^{6} - 12

y = e^{2 x}

f (x) = 3 x + 2

\frac{1}{x}

What is the normal of sqrt(1-tanh(5x)),\at x=0 ?
The normal of sqrt(1-tanh(5x)),\at x=0 is y= 2/5 x+1