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

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

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

tangent of

f (x) = sin (sin (x)), 0, 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)

Compute the slope of

f (x) = sin (sin (x))

x = π : m = - 1

Find the line with slope m=

- 1

and passing through

(π, 0) : y = - x + π

y = - x + π

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

\frac{\partial}{\partial x} (2 x^{3} + x y^{2} + 5 x^{2} + y^{2})

arccos (4 x^{2})

\int t^{3} sin (t^{2}) d t

sin (4 t + 5 π)

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