What is the tangent of y=sqrt(2x+1) ?

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

tangent of y=sqrt(2x+1)

tangent of

y = 2 x + 1

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0} + 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2 a _{0} + 1}

Find the line with slope m=

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

and passing through

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

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

in v erse l a pl a ce \frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )}

y^{^{'}} = \frac{( 5 x ^{3} + y ^{3} )}{x y ^{2}}

d er i v a t i v e f (x) = 6 x + 1

d er i v a t i v e r (x) = sin (x^{2}) + \frac{1 - x}{1 + x}

d er i v a t i v e 3 arctan (x - 1 + x^{2})

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