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

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

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

tangent of

y = sin (sin (x)), (2 π, 0)

y = x - 2 π

Solution steps

Find the slope of

y = sin (sin (x)) : \frac{d y}{d x} = cos (sin (x)) cos (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 = 1

Find the line with slope m=

1

and passing through

(2 π, 0) : y = x - 2 π

y = x - 2 π

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 6 e^{- x}

\int x sin (\frac{1}{10} x) d x

\int x + 1 ln (x + 1) d x

\int 5 x (2 + 3 x^{2})^{8} d x

\frac{10}{s ^{3} + s ^{2} + 10 s}

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