What is the tangent of 8sqrt(x)-2 ?

The tangent of 8sqrt(x)-2 is y= 4/(sqrt(a_{0))}x+4sqrt(a_{0)}-2

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tangent of 8sqrt(x)-2

tangent of

8 x - 2

y = \frac{4}{a _{0}} x + 4 a_{0} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0} - 2)

Find the slope of

y = 8 x - 2 : \frac{d y}{d x} = \frac{4}{x}

Compute the slope of

y = 8 x - 2

x = a_{0} : m = \frac{4}{a _{0}}

Find the line with slope m=

\frac{4}{a _{0}}

and passing through

(a_{0}, 8 a_{0} - 2) : y = \frac{4}{a _{0}} x + 4 a_{0} - 2

y = \frac{4}{a _{0}} x + 4 a_{0} - 2

\int_{0}^{1} (3 e^{- x} - e^{- 2 x}) d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 27 + e^{- 3 x}

\int e^{3 l n (4 x - 5)} d x

\int t sin^{2} (t) d t

f (x) = 5 - 3 x

What is the tangent of 8sqrt(x)-2 ?
The tangent of 8sqrt(x)-2 is y= 4/(sqrt(a_{0))}x+4sqrt(a_{0)}-2