What is the tangent of f(x)=sin(sin(x)),\at x=2pi ?

The tangent of f(x)=sin(sin(x)),\at x=2pi is y=x-2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sin(sin(x)),\at x=2pi

tangent of

f (x) = sin (sin (x)), at x = 2 π

y = x - 2 π

Solution steps

Find the tangent point:

(2 π, sin (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 = 2 π : m = 1

Find the line with slope m=

1

and passing through

(2 π, sin (0)) : y = x - 2 π

y = x - 2 π

x \to - 1 + lim (arcsin (x))

f (x) = ln (\frac{x + 3}{x - 3})

\frac{d}{d x} ((4 x + 5)^{2})

f (x) = sin (x^{3})

(- 4, 2), (0, 4)

What is the tangent of f(x)=sin(sin(x)),\at x=2pi ?
The tangent of f(x)=sin(sin(x)),\at x=2pi is y=x-2pi