What is the tangent of (9-t)^{1/2} ?

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

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tangent of (9-t)^{1/2}

tangent of

(9 - t)^{\frac{1}{2}}

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

Solution steps

Compute the tangent line to the general point

t = a_{0}

Find the tangent point:

(a_{0}, (9 - a_{0})^{\frac{1}{2}})

Find the slope of

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

Compute the slope of

y = (9 - t)^{\frac{1}{2}}

t = a_{0} : m = - \frac{1}{2 ( 9 - a _{0} ) ^{\frac{1}{2}}}

Find the line with slope m=

- \frac{1}{2 ( 9 - a _{0} ) ^{\frac{1}{2}}}

and passing through

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

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

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

f (x) = \frac{ln ( x )}{13 + x}

\int_{1}^{12} 2 x d x

\int \frac{9 e ^{2 t}}{1 + e ^{4 t}} d t

f (x) = \frac{1}{2} x^{2} + 7 x

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