What is the tangent of f(x)=sin(sin(x)),\at x=3pi ?

The tangent of f(x)=sin(sin(x)),\at x=3pi is y=-x+3pi

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tangent of f(x)=sin(sin(x)),\at x=3pi

tangent of

f (x) = sin (sin (x)), at x = 3 π

y = - x + 3 π

Solution steps

Find the tangent point:

(3 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 = 3 π : m = - 1

Find the line with slope m=

- 1

and passing through

(3 π, 0) : y = - x + 3 π

y = - x + 3 π

\int x^{9} e^{7 x^{5}} d x

\int x^{3} - 2 x + 2 d x

\int \frac{1 + x ^{2}}{x} d x

\int lo g_{4} (3 x + 1) d x

3 y^{^{''}} + 2 y^{^{'}} - y = 0

What is the tangent of f(x)=sin(sin(x)),\at x=3pi ?
The tangent of f(x)=sin(sin(x)),\at x=3pi is y=-x+3pi