What is the tangent of-8sin(x) ?

The tangent of-8sin(x) is y=-8cos(a_{0})x-8sin(a_{0})+8a_{0}cos(a_{0})

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tangent of-8sin(x)

tangent of

- 8 sin (x)

y = - 8 cos (a_{0}) x - 8 sin (a_{0}) + 8 a_{0} cos (a_{0})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 8 sin (a_{0}))

Find the slope of

y = - 8 sin (x) : \frac{d y}{d x} = - 8 cos (x)

Compute the slope of

y = - 8 sin (x)

x = a_{0} : m = - 8 cos (a_{0})

Find the line with slope m=

- 8 cos (a_{0})

and passing through

(a_{0}, - 8 sin (a_{0})) : y = - 8 cos (a_{0}) x - 8 sin (a_{0}) + 8 a_{0} cos (a_{0})

y = - 8 cos (a_{0}) x - 8 sin (a_{0}) + 8 a_{0} cos (a_{0})

\frac{d}{d x} (x^{4} + 2 x^{2} + 1)

\frac{d}{d t} (\frac{cos ( t )}{t})

\int_{0}^{1} x^{2} e^{- x} d x

\int (x - 4)^{3} d x

\frac{d}{d x} (\frac{tan ( x )}{e ^{x}})

What is the tangent of-8sin(x) ?
The tangent of-8sin(x) is y=-8cos(a_{0})x-8sin(a_{0})+8a_{0}cos(a_{0})