What is the tangent of-(12)/((4x+1)^2) ?

The tangent of-(12)/((4x+1)^2) is y=(96)/((4a_{0)+1)^3}x+(-144a_{0}-12)/((4a_{0)+1)^3}

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tangent of-(12)/((4x+1)^2)

tangent of

- \frac{12}{( 4 x + 1 ) ^{2}}

y = \frac{96}{( 4 a _{0} + 1 ) ^{3}} x + \frac{- 144 a _{0} - 12}{( 4 a _{0} + 1 ) ^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{12}{( 4 a _{0} + 1 ) ^{2}})

Find the slope of

y = - \frac{12}{( 4 x + 1 ) ^{2}} : \frac{d y}{d x} = \frac{96}{( 4 x + 1 ) ^{3}}

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{96}{( 4 a _{0} + 1 ) ^{3}}

Find the line with slope m=

\frac{96}{( 4 a _{0} + 1 ) ^{3}}

and passing through

(a_{0}, - \frac{12}{( 4 a _{0} + 1 ) ^{2}}) : y = \frac{96}{( 4 a _{0} + 1 ) ^{3}} x + \frac{- 144 a _{0} - 12}{( 4 a _{0} + 1 ) ^{3}}

y = \frac{96}{( 4 a _{0} + 1 ) ^{3}} x + \frac{- 144 a _{0} - 12}{( 4 a _{0} + 1 ) ^{3}}

\int \frac{1}{( e ^{y} + 1 ) ^{2} e ^{- y}} d y

\frac{d y}{d x} = \frac{4 cos ( 2 x )}{3 - sin ( 2 x )}

\int x e^{- \frac{x}{5}} d x

(\frac{1}{x ^{2} + 1})^{^{'}}

im pl i c i t \frac{d y}{d x}, y^{2} - x = 0

What is the tangent of-(12)/((4x+1)^2) ?
The tangent of-(12)/((4x+1)^2) is y=(96)/((4a_{0)+1)^3}x+(-144a_{0}-12)/((4a_{0)+1)^3}