What is the tangent of f(x)=sin(sin(x)),(4pi,0) ?

The tangent of f(x)=sin(sin(x)),(4pi,0) is y=x-4pi

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tangent of f(x)=sin(sin(x)),(4pi,0)

tangent of

f (x) = sin (sin (x)), (4 π, 0)

y = x - 4 π

Solution steps

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))

(4 π, 0) : m = 1

Find the line with slope m=

1

and passing through

(4 π, 0) : y = x - 4 π

y = x - 4 π

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

\int x^{2} sec^{2} (x) d x

\int (x + 7)^{2} d x

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

x, \frac{x ^{2}}{4}, 0, 2

What is the tangent of f(x)=sin(sin(x)),(4pi,0) ?
The tangent of f(x)=sin(sin(x)),(4pi,0) is y=x-4pi