What is the tangent of f(x)=ln(tan(2x)),\at x= pi/8 ?

The tangent of f(x)=ln(tan(2x)),\at x= pi/8 is y=4x-pi/2

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tangent of f(x)=ln(tan(2x)),\at x= pi/8

tangent of

f (x) = ln (tan (2 x)), at x = \frac{π}{8}

y = 4 x - \frac{π}{2}

Solution steps

Find the tangent point:

(\frac{π}{8}, 0)

Find the slope of

f (x) = ln (tan (2 x)) : \frac{df ( x )}{d x} = 2 sec^{2} (2 x) cot (2 x)

Compute the slope of

f (x) = ln (tan (2 x))

x = \frac{π}{8} : m = 4

Find the line with slope m=

4

and passing through

(\frac{π}{8}, 0) : y = 4 x - \frac{π}{2}

y = 4 x - \frac{π}{2}

\frac{d}{d x} (\frac{e ^{x} + 1}{e ^{2 x} - 5 e ^{x} + 6})

\frac{\partial}{\partial x} (2 x + 3 x^{2} y)

(x e^{- 3 x})^{^{'}}

\int_{0}^{2} x^{3} - 3 x^{2} + 4 d x

y^{^{'}} - \frac{5 y}{x} = 3 x^{4}, y (1) = 5

What is the tangent of f(x)=ln(tan(2x)),\at x= pi/8 ?
The tangent of f(x)=ln(tan(2x)),\at x= pi/8 is y=4x-pi/2