What is the tangent of sqrt(x-3) ?

The tangent of sqrt(x-3) is y= 1/(2sqrt(a_{0)-3)}x+sqrt(a_{0)-3}-(a_{0})/(2sqrt(a_{0)-3)}

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tangent of sqrt(x-3)

tangent of

x - 3

y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} - 3)

Find the slope of

y = x - 3 : \frac{d y}{d x} = \frac{1}{2 x - 3}

Compute the slope of

y = x - 3

x = a_{0} : m = \frac{1}{2 a _{0} - 3}

Find the line with slope m=

\frac{1}{2 a _{0} - 3}

and passing through

(a_{0}, a_{0} - 3) : y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

\int x^{22} e^{x^{23}} d x

y = (3 - x^{2}) (x^{3} - 2 x + 1)

\frac{d x}{d t} = \frac{1 - x}{k}

f (x) = sin (7 ln (x))

x y^{^{'}} = y + x^{2} sin (x), y (\frac{π}{2}) = 0

What is the tangent of sqrt(x-3) ?
The tangent of sqrt(x-3) is y= 1/(2sqrt(a_{0)-3)}x+sqrt(a_{0)-3}-(a_{0})/(2sqrt(a_{0)-3)}