What is the tangent of 7/(2sqrt(7x+1)),\at x=9 ?

The tangent of 7/(2sqrt(7x+1)),\at x=9 is y=-49/2048 x+1337/2048

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tangent of 7/(2sqrt(7x+1)),\at x=9

tangent of

\frac{7}{2 7 x + 1}, at x = 9

y = - \frac{49}{2048} x + \frac{1337}{2048}

Solution steps

Find the tangent point:

(9, \frac{7}{16})

Find the slope of

y = \frac{7}{2 7 x + 1} : \frac{d y}{d x} = - \frac{49}{4 ( 7 x + 1 ) ^{\frac{3}{2}}}

Compute the slope of

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

x = 9 : m = - \frac{49}{2048}

Find the line with slope m=

- \frac{49}{2048}

and passing through

(9, \frac{7}{16}) : y = - \frac{49}{2048} x + \frac{1337}{2048}

y = - \frac{49}{2048} x + \frac{1337}{2048}

ln (y^{2})

\frac{d}{d x} (arctan (1 - x))

\int_{2}^{3} 2 x d x

\frac{7}{4} x^{4} - \frac{6}{5} x^{3}

x \to 0 lim (\frac{e ^{x} - 1}{17 x})

What is the tangent of 7/(2sqrt(7x+1)),\at x=9 ?
The tangent of 7/(2sqrt(7x+1)),\at x=9 is y=-49/2048 x+1337/2048