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)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{n}

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{2}}

d er i v a t i v e \frac{x ^{2} + 1}{x}

\int sin (x) cos (n) x d x

y^{^{'}} = \frac{t}{1 + t ^{2}} \cdot e^{y}

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