What is the tangent of y=sin(sin(x)),(3pi,0) ?

The tangent of y=sin(sin(x)),(3pi,0) is y=-x+3pi

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tangent of y=sin(sin(x)),(3pi,0)

tangent of

y = sin (sin (x)), (3 π, 0)

y = - x + 3 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))

(3 π, 0) : m = - 1

Find the line with slope m=

- 1

and passing through

(3 π, 0) : y = - x + 3 π

y = - x + 3 π

\frac{\partial}{\partial x} (sin (x^{2} - 5 x y))

\int \frac{1}{3 x} d x

\frac{d}{d x} ((e^{(x^{2})} + 4)^{2} (4 x)^{3} x)

\int \frac{1}{sin ( x ) cos ^{3} ( x )} d x

\int_{3}^{4} \frac{( 2 x - 3 )}{6} d x

What is the tangent of y=sin(sin(x)),(3pi,0) ?
The tangent of y=sin(sin(x)),(3pi,0) is y=-x+3pi