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)

Compute the slope of

y = 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 7 x cos (2 x) d x

\int_{0}^{1} sin^{2} (2 π t) d t

\int_{0}^{2} \frac{1}{x ^{2} - 4 x + 3} d x

\frac{d ^{5}}{d x ^{5}} ((2 x + 5)^{7})

\frac{d}{d x} (A x^{2} e^{- x})

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