What is the tangent of 3/((4x+1)^2) ?

The tangent of 3/((4x+1)^2) is y=-(24)/((4a_{0)+1)^3}x+(36a_{0}+3)/((4a_{0)+1)^3}

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tangent of 3/((4x+1)^2)

tangent of

\frac{3}{( 4 x + 1 ) ^{2}}

y = - \frac{24}{( 4 a _{0} + 1 ) ^{3}} x + \frac{36 a _{0} + 3}{( 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{3}{( 4 a _{0} + 1 ) ^{2}})

Find the slope of

y = \frac{3}{( 4 x + 1 ) ^{2}} : \frac{d y}{d x} = - \frac{24}{( 4 x + 1 ) ^{3}}

Compute the slope of

y = \frac{3}{( 4 x + 1 ) ^{2}}

x = a_{0} : m = - \frac{24}{( 4 a _{0} + 1 ) ^{3}}

Find the line with slope m=

- \frac{24}{( 4 a _{0} + 1 ) ^{3}}

and passing through

(a_{0}, \frac{3}{( 4 a _{0} + 1 ) ^{2}}) : y = - \frac{24}{( 4 a _{0} + 1 ) ^{3}} x + \frac{36 a _{0} + 3}{( 4 a _{0} + 1 ) ^{3}}

y = - \frac{24}{( 4 a _{0} + 1 ) ^{3}} x + \frac{36 a _{0} + 3}{( 4 a _{0} + 1 ) ^{3}}

\frac{d}{d x} (\frac{x ^{3}}{3} + 1)

t \to 0 lim (\frac{t}{t ^{2} - t})

\frac{d}{d x} (x + 1 - e^{x})

\int 3 sin (x) d x

What is the tangent of 3/((4x+1)^2) ?
The tangent of 3/((4x+1)^2) is y=-(24)/((4a_{0)+1)^3}x+(36a_{0}+3)/((4a_{0)+1)^3}