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

The tangent of f(x)= 3/(2sqrt(3x+1)),\at x=8 is y=-9/500 x+111/250

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tangent of f(x)= 3/(2sqrt(3x+1)),\at x=8

tangent of

f (x) = \frac{3}{2 3 x + 1}, at x = 8

y = - \frac{9}{500} x + \frac{111}{250}

Solution steps

Find the tangent point:

(8, \frac{3}{10})

Find the slope of

f (x) = \frac{3}{2 3 x + 1} : \frac{df ( x )}{d x} = - \frac{9}{4 ( 3 x + 1 ) ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{3}{2 3 x + 1}

x = 8 : m = - \frac{9}{500}

Find the line with slope m=

- \frac{9}{500}

and passing through

(8, \frac{3}{10}) : y = - \frac{9}{500} x + \frac{111}{250}

y = - \frac{9}{500} x + \frac{111}{250}

y^{^{'}} = \frac{y}{x} + \frac{x}{2 ( y + x )}

\int_{2}^{5} \frac{x}{x - 1} d x

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

\frac{\partial}{\partial x} (\frac{( 2 x + 3 y )}{4 x + 5 y})

f (x) = x^{2} + 5

What is the tangent of f(x)= 3/(2sqrt(3x+1)),\at x=8 ?
The tangent of f(x)= 3/(2sqrt(3x+1)),\at x=8 is y=-9/500 x+111/250