What is the tangent of f(x)=-sin(-8x-3)+5 ?

The tangent of f(x)=-sin(-8x-3)+5 is y=8cos(8a_{0}+3)x-sin(-8a_{0}-3)+5-8cos(8a_{0}+3)a_{0}

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tangent of f(x)=-sin(-8x-3)+5

tangent of

f (x) = - sin (- 8 x - 3) + 5

y = 8 cos (8 a_{0} + 3) x - sin (- 8 a_{0} - 3) + 5 - 8 cos (8 a_{0} + 3) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - sin (- 8 a_{0} - 3) + 5)

Find the slope of

f (x) = - sin (- 8 x - 3) + 5 : \frac{df ( x )}{d x} = 8 cos (8 x + 3)

Compute the slope of

f (x) = - sin (- 8 x - 3) + 5

x = a_{0} : m = 8 cos (8 a_{0} + 3)

Find the line with slope m=

8 cos (8 a_{0} + 3)

and passing through

(a_{0}, - sin (- 8 a_{0} - 3) + 5) : y = 8 cos (8 a_{0} + 3) x - sin (- 8 a_{0} - 3) + 5 - 8 cos (8 a_{0} + 3) a_{0}

y = 8 cos (8 a_{0} + 3) x - sin (- 8 a_{0} - 3) + 5 - 8 cos (8 a_{0} + 3) a_{0}

\int_{0}^{1} 6 x 5^{x} d x

x \to a lim (1)

2 x^{3} (2.16)

\frac{d}{d x} (120)

x \to - 0 lim (\frac{1}{x})

What is the tangent of f(x)=-sin(-8x-3)+5 ?
The tangent of f(x)=-sin(-8x-3)+5 is y=8cos(8a_{0}+3)x-sin(-8a_{0}-3)+5-8cos(8a_{0}+3)a_{0}