What is the tangent of f(x)= 1/(sqrt(2-3x)) ?

The tangent of f(x)= 1/(sqrt(2-3x)) is y= 3/(2(2-3a_{0))^{3/2}}x+(-9a_{0}+4)/(2(-3a_{0)+2)^{3/2}}

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tangent of f(x)= 1/(sqrt(2-3x))

tangent of

f (x) = \frac{1}{2 - 3 x}

y = \frac{3}{2 ( 2 - 3 a _{0} ) ^{\frac{3}{2}}} x + \frac{- 9 a _{0} + 4}{2 ( - 3 a _{0} + 2 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 - 3 a _{0}})

Find the slope of

f (x) = \frac{1}{2 - 3 x} : \frac{df ( x )}{d x} = \frac{3}{2 ( 2 - 3 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{3}{2 ( 2 - 3 a _{0} ) ^{\frac{3}{2}}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{1}{2 - 3 a _{0}}) : y = \frac{3}{2 ( 2 - 3 a _{0} ) ^{\frac{3}{2}}} x + \frac{- 9 a _{0} + 4}{2 ( - 3 a _{0} + 2 ) ^{\frac{3}{2}}}

y = \frac{3}{2 ( 2 - 3 a _{0} ) ^{\frac{3}{2}}} x + \frac{- 9 a _{0} + 4}{2 ( - 3 a _{0} + 2 ) ^{\frac{3}{2}}}

\frac{d}{d x} ((\frac{x ^{2} - 9}{4 x + 5})^{4})

\frac{1}{x} \frac{d y}{d x} - \frac{2}{x ^{2}} y = x^{2} e^{- x}

\frac{5}{x ^{5}} + \frac{3}{x ^{2}}

\int \frac{y - y}{y ^{2}} d y

\int (ln (x))^{2} d x

What is the tangent of f(x)= 1/(sqrt(2-3x)) ?
The tangent of f(x)= 1/(sqrt(2-3x)) is y= 3/(2(2-3a_{0))^{3/2}}x+(-9a_{0}+4)/(2(-3a_{0)+2)^{3/2}}