What is the tangent of y=sin(2x)+sin(2)(2x) ?

The tangent of y=sin(2x)+sin(2)(2x) is y=(cos(2a_{0})*2+2sin(2))x-2a_{0}cos(2a_{0})+sin(2a_{0})

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tangent of y=sin(2x)+sin(2)(2x)

tangent of

y = sin (2 x) + sin (2) (2 x)

y = (cos (2 a_{0}) \cdot 2 + 2 sin (2)) x - 2 a_{0} cos (2 a_{0}) + sin (2 a_{0})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, sin (2 a_{0}) + sin (2) \cdot 2 a_{0})

Find the slope of

y = sin (2 x) + sin (2) 2 x : \frac{d y}{d x} = cos (2 x) \cdot 2 + 2 sin (2)

Compute the slope of

y = sin (2 x) + sin (2) 2 x

x = a_{0} : m = cos (2 a_{0}) \cdot 2 + 2 sin (2)

Find the line with slope m=

cos (2 a_{0}) 2 + 2 sin (2)

and passing through

(a_{0}, sin (2 a_{0}) + sin (2) 2 a_{0}) : y = (cos (2 a_{0}) \cdot 2 + 2 sin (2)) x - 2 a_{0} cos (2 a_{0}) + sin (2 a_{0})

y = (cos (2 a_{0}) \cdot 2 + 2 sin (2)) x - 2 a_{0} cos (2 a_{0}) + sin (2 a_{0})

x \to - \frac{π}{2} lim (2 tan (x))

\frac{d}{d x} (\frac{e ^{x + 1}}{x ^{2} + 1})

\frac{d}{d t} (- 2 sin (t))

\int x \cdot arctan (x^{2}) d x

(10.8) (1.3)

What is the tangent of y=sin(2x)+sin(2)(2x) ?
The tangent of y=sin(2x)+sin(2)(2x) is y=(cos(2a_{0})*2+2sin(2))x-2a_{0}cos(2a_{0})+sin(2a_{0})