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

The tangent of sqrt(7x+1),\at x=9 is y= 7/16 x+65/16

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

tangent of

7 x + 1, at x = 9

y = \frac{7}{16} x + \frac{65}{16}

Solution steps

Find the tangent point:

(9, 8)

Find the slope of

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

Compute the slope of

y = 7 x + 1

x = 9 : m = \frac{7}{16}

Find the line with slope m=

\frac{7}{16}

and passing through

(9, 8) : y = \frac{7}{16} x + \frac{65}{16}

y = \frac{7}{16} x + \frac{65}{16}

cos (t)

x \to 0 lim (1 - sin^{2} (x))

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

\int \frac{( x - 2 )}{x} d x

x \to \infty lim (x \cdot (- 1))

What is the tangent of sqrt(7x+1),\at x=9 ?
The tangent of sqrt(7x+1),\at x=9 is y= 7/16 x+65/16