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

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

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

tangent of

f (x) = sin (sin (x)), at x = 2 π, 0

y = x - 2 π

Solution steps

Find the tangent point:

(2 π, sin (0))

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))

x = 2 π : m = 1

Find the line with slope m=

1

and passing through

(2 π, sin (0)) : y = x - 2 π

y = x - 2 π

\frac{d}{d x} (3 sin (x) + 2 cos (x))

\int 6 x^{6} d x

\frac{d}{d x} (\frac{e ^{x}}{4 - x})

y^{^{''}} - 2 y^{^{'}} + y = t e^{t}

\int p (p + 5)^{5} d p

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