What is the tangent of f(x)=sin(sin(x)),(2pi,0) ?

The tangent of f(x)=sin(sin(x)),(2pi,0) is y=x-2pi

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tangent of f(x)=sin(sin(x)),(2pi,0)

tangent of

f (x) = sin (sin (x)), (2 π, 0)

y = x - 2 π

Solution steps

Find the slope of

f (x) = sin (sin (x)) : \frac{df ( x )}{d x} = cos (sin (x)) cos (x)

Compute the slope of

f (x) = sin (sin (x))

(2 π, 0) : m = 1

Find the line with slope m=

1

and passing through

(2 π, 0) : y = x - 2 π

y = x - 2 π

\int \frac{I n x}{x} d x

\int \frac{12}{sec ( 2 x ) - 1} d x

\int \frac{1}{x 1 + 2 x} d x

y^{^{'}} - y = e^{- t} sin (t)

n = 1 \sum \infty ln (n^{2} + 2 n)

What is the tangent of f(x)=sin(sin(x)),(2pi,0) ?
The tangent of f(x)=sin(sin(x)),(2pi,0) is y=x-2pi