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

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

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

tangent of

3 x + 1, at x = 5

y = \frac{3}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(5, 4)

Find the slope of

y = 3 x + 1 : \frac{d y}{d x} = \frac{3}{2 3 x + 1}

Compute the slope of

y = 3 x + 1

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

Find the line with slope m=

\frac{3}{8}

and passing through

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

y = \frac{3}{8} x + \frac{17}{8}

\frac{\partial}{\partial x} (e^{x} - 1)

\frac{d}{d x} (\frac{1}{5 x ^{\frac{4}{5}}})

\int w d w

\int x^{3} + 2 x d x

\frac{d y}{d x} = 1 - x - y

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