What is the tangent of 7/(2x+1),\at a=4 ?

The tangent of 7/(2x+1),\at a=4 is y=-(14)/((2a+1)^2)x+(28a+7)/((2a+1)^2)

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tangent of 7/(2x+1),\at a=4

tangent of

\frac{7}{2 x + 1}, at a = 4

y = - \frac{14}{( 2 a + 1 ) ^{2}} x + \frac{28 a + 7}{( 2 a + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, \frac{7}{2 a + 1})

Find the slope of

y = \frac{7}{2 x + 1} : \frac{d y}{d x} = - \frac{14}{( 2 x + 1 ) ^{2}}

Compute the slope of

y = \frac{7}{2 x + 1}

x = a : m = - \frac{14}{( 2 a + 1 ) ^{2}}

Find the line with slope m=

- \frac{14}{( 2 a + 1 ) ^{2}}

and passing through

(a, \frac{7}{2 a + 1}) : y = - \frac{14}{( 2 a + 1 ) ^{2}} x + \frac{28 a + 7}{( 2 a + 1 ) ^{2}}

y = - \frac{14}{( 2 a + 1 ) ^{2}} x + \frac{28 a + 7}{( 2 a + 1 ) ^{2}}

2 x^{3} - 6 x^{2} - 90 x + 1

4 x - 4 x

f (x) = e^{8 x}

x \to \frac{x}{4} lim (tan (2 x))

\int cos (2 x) e^{3 x} d x

What is the tangent of 7/(2x+1),\at a=4 ?
The tangent of 7/(2x+1),\at a=4 is y=-(14)/((2a+1)^2)x+(28a+7)/((2a+1)^2)