What is the tangent of (6x)/((3-2x)^3) ?

The tangent of (6x)/((3-2x)^3) is y=(6(3+4a_{0}))/((3-2a_{0))^4}x-(36a_{0}^2)/((-2a_{0)+3)^4}

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tangent of (6x)/((3-2x)^3)

tangent of

\frac{6 x}{( 3 - 2 x ) ^{3}}

y = \frac{6 ( 3 + 4 a _{0} )}{( 3 - 2 a _{0} ) ^{4}} x - \frac{36 a _{0}^{2}}{( - 2 a _{0} + 3 ) ^{4}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6 a _{0}}{( 3 - 2 a _{0} ) ^{3}})

Find the slope of

y = \frac{6 x}{( 3 - 2 x ) ^{3}} : \frac{d y}{d x} = \frac{6 ( 3 + 4 x )}{( 3 - 2 x ) ^{4}}

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{6 ( 3 + 4 a _{0} )}{( 3 - 2 a _{0} ) ^{4}}

Find the line with slope m=

\frac{6 ( 3 + 4 a _{0} )}{( 3 - 2 a _{0} ) ^{4}}

and passing through

(a_{0}, \frac{6 a _{0}}{( 3 - 2 a _{0} ) ^{3}}) : y = \frac{6 ( 3 + 4 a _{0} )}{( 3 - 2 a _{0} ) ^{4}} x - \frac{36 a _{0}^{2}}{( - 2 a _{0} + 3 ) ^{4}}

y = \frac{6 ( 3 + 4 a _{0} )}{( 3 - 2 a _{0} ) ^{4}} x - \frac{36 a _{0}^{2}}{( - 2 a _{0} + 3 ) ^{4}}

\frac{d}{d x} ((6 x + 5)^{2})

y^{^{''}} + 16 y = 24 sin (4 t)

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

f (x) = - 3 x

\int (sin (x) - cos (x))^{2} d x

What is the tangent of (6x)/((3-2x)^3) ?
The tangent of (6x)/((3-2x)^3) is y=(6(3+4a_{0}))/((3-2a_{0))^4}x-(36a_{0}^2)/((-2a_{0)+3)^4}