What is the tangent of 3/(2sqrt(3x+7)),\at x=6 ?

The tangent of 3/(2sqrt(3x+7)),\at x=6 is y=-9/500 x+51/125

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tangent of 3/(2sqrt(3x+7)),\at x=6

tangent of

\frac{3}{2 3 x + 7}, at x = 6

y = - \frac{9}{500} x + \frac{51}{125}

Solution steps

Find the tangent point:

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

Find the slope of

y = \frac{3}{2 3 x + 7} : \frac{d y}{d x} = - \frac{9}{4 ( 3 x + 7 ) ^{\frac{3}{2}}}

Compute the slope of

y = \frac{3}{2 3 x + 7}

x = 6 : m = - \frac{9}{500}

Find the line with slope m=

- \frac{9}{500}

and passing through

(6, \frac{3}{10}) : y = - \frac{9}{500} x + \frac{51}{125}

y = - \frac{9}{500} x + \frac{51}{125}

\frac{d}{d x} (x^{2} arctan (x))

n = 0 \sum \infty \frac{1}{3 ^{n} + 1}

\frac{1}{2} x^{4} + 11 x^{3} + 30 x^{2}

\frac{2 d y}{d x} + 5 y = 4

x \to 0 lim (\frac{1}{3})

What is the tangent of 3/(2sqrt(3x+7)),\at x=6 ?
The tangent of 3/(2sqrt(3x+7)),\at x=6 is y=-9/500 x+51/125