What is the tangent of sqrt(3x+1),\at a=8 ?

The tangent of sqrt(3x+1),\at a=8 is y= 3/(2sqrt(3a+1))x+sqrt(3a+1)-(3a)/(2sqrt(3a+1))

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tangent of sqrt(3x+1),\at a=8

tangent of

3 x + 1, at a = 8

y = \frac{3}{2 3 a + 1} x + 3 a + 1 - \frac{3 a}{2 3 a + 1}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, 3 a + 1)

Find the slope of

y = 3 x + 1 : \frac{d y}{d x} = \frac{3}{2 3 x + 1}

Compute the slope of

y = 3 x + 1

x = a : m = \frac{3}{2 3 a + 1}

Find the line with slope m=

\frac{3}{2 3 a + 1}

and passing through

(a, 3 a + 1) : y = \frac{3}{2 3 a + 1} x + 3 a + 1 - \frac{3 a}{2 3 a + 1}

y = \frac{3}{2 3 a + 1} x + 3 a + 1 - \frac{3 a}{2 3 a + 1}

\int sin (\frac{1 - x}{3}) d x

cos^{2} (ln (e^{2 x + 1} (- 2 x)))

\frac{3 e ^{- π s}}{s ^{2} + 9}

f (x) = 4 x^{2} - 1, (- 6, 3)

y^{^{'}} = (x + y - 2)^{2}, y (0) = 0

What is the tangent of sqrt(3x+1),\at a=8 ?
The tangent of sqrt(3x+1),\at a=8 is y= 3/(2sqrt(3a+1))x+sqrt(3a+1)-(3a)/(2sqrt(3a+1))