What is the tangent of sqrt(9-x) ?

The tangent of sqrt(9-x) is y=-1/(2sqrt(9-a_{0))}x+sqrt(9-a_{0)}+(a_{0})/(2sqrt(9-a_{0))}

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tangent of sqrt(9-x)

tangent of

9 - x

y = - \frac{1}{2 9 - a _{0}} x + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 - a_{0})

Find the slope of

y = 9 - x : \frac{d y}{d x} = - \frac{1}{2 9 - x}

Compute the slope of

y = 9 - x

x = a_{0} : m = - \frac{1}{2 9 - a _{0}}

Find the line with slope m=

- \frac{1}{2 9 - a _{0}}

and passing through

(a_{0}, 9 - a_{0}) : y = - \frac{1}{2 9 - a _{0}} x + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

y = - \frac{1}{2 9 - a _{0}} x + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

tan^{2} (8 θ)

\frac{d}{d x} (2 ln (1 + x^{2}))

\frac{\partial}{\partial x} (ln (x^{3} + y^{3}))

\frac{\partial}{\partial x} (B (\frac{G ( B , x )}{a})^{\frac{1}{x}})

\frac{\partial}{\partial x} (- 2 x^{3} + sin (x y) + 4 e^{4 y})

What is the tangent of sqrt(9-x) ?
The tangent of sqrt(9-x) is y=-1/(2sqrt(9-a_{0))}x+sqrt(9-a_{0)}+(a_{0})/(2sqrt(9-a_{0))}