What is the tangent of 8cos(x) ?

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

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tangent of 8cos(x)

tangent of

8 cos (x)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 cos (a_{0}))

Find the slope of

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

Compute the slope of

y = 8 cos (x)

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

Find the line with slope m=

- 8 sin (a_{0})

and passing through

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

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

\int \frac{y}{y ^{2} + 1} d y

(x + y) d x + (x + y - 4) d y = 0

\int x^{- 1} e^{x} d x

\frac{d}{d x} (\frac{8 x}{x + 5})

(3 x^{2} - 4 x) e^{x}

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