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

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

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tangent of f(x)=sin(sin(x)),(pi,0)

tangent of

f (x) = sin (sin (x)), (π, 0)

y = - x + π

Solution steps

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))

(π, 0) : m = - 1

Find the line with slope m=

- 1

and passing through

(π, 0) : y = - x + π

y = - x + π

\int \frac{9 x ( 6 x ^{2} + 5 ) ^{\frac{1}{2}}}{2} d x

\int \frac{1}{19 y} d y

\frac{4}{s ^{2} + 7 s + 12}

x \to 0 lim (e^{x} + 1)

a + b e^{θ}

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