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

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

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

tangent of

f (x) = sin (sin (x)), at x = π

y = - x + π

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

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

(π, 0) : y = - x + π

y = - x + π

x \to 0 lim (\frac{tan ( 3 x ) - sin ( 3 x )}{x ^{3}})

\frac{d}{d x} ((1 - x^{2})^{2})

\frac{d y}{d x} = \frac{y + 3}{x + 3}

\frac{x ^{3}}{6 x - 3}

3 x^{2} + 4, [1, 2]

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