What is the tangent of f(x)=sin(sin(x)),\at x=4pi ?

The tangent of f(x)=sin(sin(x)),\at x=4pi is y=x-4pi

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tangent of f(x)=sin(sin(x)),\at x=4pi

tangent of

f (x) = sin (sin (x)), at x = 4 π

y = x - 4 π

Solution steps

Find the tangent point:

(4 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 = 4 π : m = 1

Find the line with slope m=

1

and passing through

(4 π, sin (0)) : y = x - 4 π

y = x - 4 π

y^{^{'}} = y (1 - y)

y = e^{4 x^{2}} + 1

f (x) = tan (x), at x = 1

19 x, x^{2}

n = 0 \sum \infty \frac{1}{3} (\frac{2}{3})^{n}

What is the tangent of f(x)=sin(sin(x)),\at x=4pi ?
The tangent of f(x)=sin(sin(x)),\at x=4pi is y=x-4pi