What is the tangent of y=sqrt(2x+1),\at x= 1/2 ?

The tangent of y=sqrt(2x+1),\at x= 1/2 is y= 1/(sqrt(2))x+sqrt(2)-1/(2sqrt(2))

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tangent of y=sqrt(2x+1),\at x= 1/2

tangent of

y = 2 x + 1, at x = \frac{1}{2}

y = \frac{1}{2} x + 2 - \frac{1}{2 2}

Solution steps

Find the tangent point:

(\frac{1}{2}, 2)

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}

Find the line with slope m=

\frac{1}{2}

and passing through

(\frac{1}{2}, 2) : y = \frac{1}{2} x + 2 - \frac{1}{2 2}

y = \frac{1}{2} x + 2 - \frac{1}{2 2}

\frac{\partial}{\partial x} (\frac{2}{( x - 3 t ) ^{2} + 1})

\int (x - 4) d x

\frac{d}{d θ} (\frac{5}{2 cos ( θ ) + 7 sin ( θ )})

t an g e n t f (x) = 3 x^{4} - 8 x^{3} + 2

\frac{d}{d x} (5 e^{- 2 x} sin (3 x))

What is the tangent of y=sqrt(2x+1),\at x= 1/2 ?
The tangent of y=sqrt(2x+1),\at x= 1/2 is y= 1/(sqrt(2))x+sqrt(2)-1/(2sqrt(2))