What is the tangent of (3+sin(2x))^2,\at x=0 ?

The tangent of (3+sin(2x))^2,\at x=0 is y=12x+9

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tangent of (3+sin(2x))^2,\at x=0

tangent of

(3 + sin (2 x))^{2}, at x = 0

y = 12 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

y = (3 + sin (2 x))^{2} : \frac{d y}{d x} = 4 cos (2 x) (3 + sin (2 x))

Compute the slope of

y = (3 + sin (2 x))^{2}

x = 0 : m = 12

Find the line with slope m=

12

and passing through

(0, 9) : y = 12 x + 9

y = 12 x + 9

\frac{2}{x ^{2} - x}

\int 2 x + 8 x^{- 2} d x

f (x) = x^{2} + 10 x, at x = 8

x \to \infty lim (x^{2} e^{2 x})

x \to - 2 lim (\frac{x ^{3} + 1}{x + 1})

What is the tangent of (3+sin(2x))^2,\at x=0 ?
The tangent of (3+sin(2x))^2,\at x=0 is y=12x+9