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} (\frac{8 x ^{2}}{2 x - 10})

\frac{d}{d x} (\frac{x}{t})

5 \to 1 lim (\frac{\frac{1}{5}}{5 - 1})

y = \frac{8}{x}, y = 2 x, x = 4

\frac{d}{d x} ((3 x^{2} - 5 x)^{9} (- 2 x^{4} + 2 x - 8)^{4})

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