What is the tangent of sqrt(5x+1),\at x=3 ?

The tangent of sqrt(5x+1),\at x=3 is y= 5/8 x+17/8

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tangent of sqrt(5x+1),\at x=3

tangent of

5 x + 1, at x = 3

y = \frac{5}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

y = 5 x + 1 : \frac{d y}{d x} = \frac{5}{2 5 x + 1}

Compute the slope of

y = 5 x + 1

x = 3 : m = \frac{5}{8}

Find the line with slope m=

\frac{5}{8}

and passing through

(3, 4) : y = \frac{5}{8} x + \frac{17}{8}

y = \frac{5}{8} x + \frac{17}{8}

\frac{d}{d x} - \frac{3 x ^{2} sin ( \frac{x ^{3}}{4} )}{4}

6 sec (x), at x = \frac{π}{6}

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

x \to - 2 lim (f (x) x)

\frac{d}{d x} (cos (x (x^{2} + 1)) + \frac{3}{2})

What is the tangent of sqrt(5x+1),\at x=3 ?
The tangent of sqrt(5x+1),\at x=3 is y= 5/8 x+17/8