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

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

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

tangent of

y = sin (sin (x)), (π, 0)

y = - x + π

Solution steps

Find the slope of

y = sin (sin (x)) : \frac{d y}{d x} = cos (sin (x)) cos (x)

Compute the slope of

y = 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

\frac{d}{d x} (sin (x^{2} - 4))

\frac{\partial}{\partial u} ((u^{2} v - v^{3})^{5})

\int x y sin (z) d z

y = x^{- \frac{1}{7}}, (1, 1)

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 6 x + 6 y)

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