What is the tangent of sqrt(4x+20) ?

The tangent of sqrt(4x+20) is y= 1/(sqrt(a_{0)+5)}x+2sqrt(a_{0)+5}-(a_{0})/(sqrt(a_{0)+5)}

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tangent of sqrt(4x+20)

tangent of

4 x + 20

y = \frac{1}{a _{0} + 5} x + 2 a_{0} + 5 - \frac{a _{0}}{a _{0} + 5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0} + 5)

Find the slope of

y = 4 x + 20 : \frac{d y}{d x} = \frac{1}{x + 5}

Compute the slope of

y = 4 x + 20

x = a_{0} : m = \frac{1}{a _{0} + 5}

Find the line with slope m=

\frac{1}{a _{0} + 5}

and passing through

(a_{0}, 2 a_{0} + 5) : y = \frac{1}{a _{0} + 5} x + 2 a_{0} + 5 - \frac{a _{0}}{a _{0} + 5}

y = \frac{1}{a _{0} + 5} x + 2 a_{0} + 5 - \frac{a _{0}}{a _{0} + 5}

f (x) = \frac{x + 3}{2 x + 1}, at x = - 1

2 x^{- \frac{1}{4}}

x \to \infty lim (9^{x})

\frac{d}{d x} (\frac{1}{1 + ln ( x )})

y^{^{'}} - \frac{y}{x} = \frac{x y}{x}

What is the tangent of sqrt(4x+20) ?
The tangent of sqrt(4x+20) is y= 1/(sqrt(a_{0)+5)}x+2sqrt(a_{0)+5}-(a_{0})/(sqrt(a_{0)+5)}