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

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

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

tangent of

y = sin (2 x), at x = \frac{π}{2}

y = - 2 x + π

Solution steps

Find the tangent point:

(\frac{π}{2}, 0)

Find the slope of

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

Compute the slope of

y = sin (2 x)

x = \frac{π}{2} : m = - 2

Find the line with slope m=

- 2

and passing through

(\frac{π}{2}, 0) : y = - 2 x + π

y = - 2 x + π

2 x^{2} + 3 x - 2, at x = 0

f (x) = \frac{3}{x ^{2}} + \frac{5}{x ^{4}}

\frac{d y}{d x} = 4 x + \frac{4 x}{16 - x ^{2}}

x \to 8 lim (\frac{3 x ^{2}}{4})

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

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