What is the tangent of sqrt(x+8) ?

The tangent of sqrt(x+8) is y= 1/(2sqrt(a_{0)+8)}x+sqrt(a_{0)+8}-(a_{0})/(2sqrt(a_{0)+8)}

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tangent of sqrt(x+8)

tangent of

x + 8

y = \frac{1}{2 a _{0} + 8} x + a_{0} + 8 - \frac{a _{0}}{2 a _{0} + 8}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x + 8 : \frac{d y}{d x} = \frac{1}{2 x + 8}

Compute the slope of

y = x + 8

x = a_{0} : m = \frac{1}{2 a _{0} + 8}

Find the line with slope m=

\frac{1}{2 a _{0} + 8}

and passing through

(a_{0}, a_{0} + 8) : y = \frac{1}{2 a _{0} + 8} x + a_{0} + 8 - \frac{a _{0}}{2 a _{0} + 8}

y = \frac{1}{2 a _{0} + 8} x + a_{0} + 8 - \frac{a _{0}}{2 a _{0} + 8}

n = 0 \sum \infty 3^{n} 7^{- n}

\frac{\partial}{\partial y} (- x y^{2})

y = 2 x e^{- 6 x} - 4 e^{x^{3}}

f (x) = \frac{10 x}{x ^{2} - 6}, (4, 4)

4 x^{2} - 3

What is the tangent of sqrt(x+8) ?
The tangent of sqrt(x+8) is y= 1/(2sqrt(a_{0)+8)}x+sqrt(a_{0)+8}-(a_{0})/(2sqrt(a_{0)+8)}