What is the tangent of sqrt(7x+1) ?

The tangent of sqrt(7x+1) is y= 7/(2sqrt(7a_{0)+1)}x+sqrt(7a_{0)+1}-(7a_{0})/(2sqrt(7a_{0)+1)}

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tangent of sqrt(7x+1)

tangent of

7 x + 1

y = \frac{7}{2 7 a _{0} + 1} x + 7 a_{0} + 1 - \frac{7 a _{0}}{2 7 a _{0} + 1}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 a_{0} + 1)

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 = a_{0} : m = \frac{7}{2 7 a _{0} + 1}

Find the line with slope m=

\frac{7}{2 7 a _{0} + 1}

and passing through

(a_{0}, 7 a_{0} + 1) : y = \frac{7}{2 7 a _{0} + 1} x + 7 a_{0} + 1 - \frac{7 a _{0}}{2 7 a _{0} + 1}

y = \frac{7}{2 7 a _{0} + 1} x + 7 a_{0} + 1 - \frac{7 a _{0}}{2 7 a _{0} + 1}

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What is the tangent of sqrt(7x+1) ?
The tangent of sqrt(7x+1) is y= 7/(2sqrt(7a_{0)+1)}x+sqrt(7a_{0)+1}-(7a_{0})/(2sqrt(7a_{0)+1)}