What is the tangent of 8x+7cos(x) ?

The tangent of 8x+7cos(x) is y=(8-7sin(a_{0}))x+7a_{0}sin(a_{0})+7cos(a_{0})

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tangent of 8x+7cos(x)

tangent of

8 x + 7 cos (x)

y = (8 - 7 sin (a_{0})) x + 7 a_{0} sin (a_{0}) + 7 cos (a_{0})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0} + 7 cos (a_{0}))

Find the slope of

y = 8 x + 7 cos (x) : \frac{d y}{d x} = 8 - 7 sin (x)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8 - 7 sin (a_{0})

Find the line with slope m=

8 - 7 sin (a_{0})

and passing through

(a_{0}, 8 a_{0} + 7 cos (a_{0})) : y = (8 - 7 sin (a_{0})) x + 7 a_{0} sin (a_{0}) + 7 cos (a_{0})

y = (8 - 7 sin (a_{0})) x + 7 a_{0} sin (a_{0}) + 7 cos (a_{0})

f (x) = 2 x^{2} - 5 x + 3, at x = 2

\frac{d}{d t} (\frac{8 t}{8 + t})

\int \frac{1}{t ^{2} - 8 t + 20} d t

x \to 0 lim (\frac{1}{cos ( 3 x )})

y^{^{''}} + 7 y^{^{'}} - 8 y = 0

What is the tangent of 8x+7cos(x) ?
The tangent of 8x+7cos(x) is y=(8-7sin(a_{0}))x+7a_{0}sin(a_{0})+7cos(a_{0})