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

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

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tangent of sqrt(9-t)

tangent of

9 - t

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

Solution steps

Compute the tangent line to the general point

t = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 9 - t

t = 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}} t + 9 - a_{0} + \frac{a _{0}}{2 9 - a _{0}}

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

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 3 x}{x + 1})

n = 1 \sum \infty 6 e^{- 5 n}

\int \frac{12 x ^{3} - 24 x ^{2} + 5}{x ^{2} - 2 x} d x

\frac{\partial}{\partial x} (x + 4 y)

\int e^{2 x} 2 x d x

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