What is the tangent of y= 1/(sqrt(4-x)) ?

The tangent of y= 1/(sqrt(4-x)) is y= 1/(2(4-a_{0))^{3/2}}x+(-3a_{0}+8)/(2(-a_{0)+4)^{3/2}}

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tangent of y= 1/(sqrt(4-x))

tangent of

y = \frac{1}{4 - x}

y = \frac{1}{2 ( 4 - a _{0} ) ^{\frac{3}{2}}} x + \frac{- 3 a _{0} + 8}{2 ( - a _{0} + 4 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{4 - a _{0}})

Find the slope of

y = \frac{1}{4 - x} : \frac{d y}{d x} = \frac{1}{2 ( 4 - x ) ^{\frac{3}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2 ( 4 - a _{0} ) ^{\frac{3}{2}}}

Find the line with slope m=

\frac{1}{2 ( 4 - a _{0} ) ^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{4 - a _{0}}) : y = \frac{1}{2 ( 4 - a _{0} ) ^{\frac{3}{2}}} x + \frac{- 3 a _{0} + 8}{2 ( - a _{0} + 4 ) ^{\frac{3}{2}}}

y = \frac{1}{2 ( 4 - a _{0} ) ^{\frac{3}{2}}} x + \frac{- 3 a _{0} + 8}{2 ( - a _{0} + 4 ) ^{\frac{3}{2}}}

\int \frac{( 4 x - 7 )}{x ^{2} - 6 x + 13} d x

- \frac{7}{x ^{2}}

y = 8 - x^{2}, y = x^{2}, x = - 3, x = 3

(x^{2} + 4) d y = (2 x - 8 x y) d x

\int - \frac{2}{πn} cos (\frac{πn}{2} x) d x

What is the tangent of y= 1/(sqrt(4-x)) ?
The tangent of y= 1/(sqrt(4-x)) is y= 1/(2(4-a_{0))^{3/2}}x+(-3a_{0}+8)/(2(-a_{0)+4)^{3/2}}