What is the tangent of y=sqrt(5x+1) ?

The tangent of y=sqrt(5x+1) is y= 5/(2sqrt(5a_{0)+1)}x+sqrt(5a_{0)+1}-(5a_{0})/(2sqrt(5a_{0)+1)}

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tangent of y=sqrt(5x+1)

tangent of

y = 5 x + 1

y = \frac{5}{2 5 a _{0} + 1} x + 5 a_{0} + 1 - \frac{5 a _{0}}{2 5 a _{0} + 1}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0} + 1)

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 = a_{0} : m = \frac{5}{2 5 a _{0} + 1}

Find the line with slope m=

\frac{5}{2 5 a _{0} + 1}

and passing through

(a_{0}, 5 a_{0} + 1) : y = \frac{5}{2 5 a _{0} + 1} x + 5 a_{0} + 1 - \frac{5 a _{0}}{2 5 a _{0} + 1}

y = \frac{5}{2 5 a _{0} + 1} x + 5 a_{0} + 1 - \frac{5 a _{0}}{2 5 a _{0} + 1}

n = 0 \sum \infty (\frac{x}{5})^{n}

\int \frac{x}{x ^{2} + y ^{2} + z ^{2}} d x

x \to 0 lim (\frac{sin ( 12 x )}{x})

\frac{d}{d x} (\frac{x - 1}{x} + ln (x))

4 sin (4 t)

What is the tangent of y=sqrt(5x+1) ?
The tangent of y=sqrt(5x+1) is y= 5/(2sqrt(5a_{0)+1)}x+sqrt(5a_{0)+1}-(5a_{0})/(2sqrt(5a_{0)+1)}