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

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

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tangent of 1/(4x+3)

tangent of

\frac{1}{4 x + 3}

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

Find the slope of

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

Compute the slope of

y = \frac{1}{4 x + 3}

x = a_{0} : m = - \frac{4}{( 4 a _{0} + 3 ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

\frac{\partial}{\partial x} (7 x - 3 y + 4)

\int_{0}^{1} x^{6} - x d x

y^{^{'}} + y \cdot cos (x) = sin (x) \cdot cos (x)

\int \frac{e ^{x} + sec ^{2} ( x )}{e ^{x} + tan ( x )} d x

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