What is the tangent of y=sqrt(2x+1)(4.3) ?

The tangent of y=sqrt(2x+1)(4.3) is y=(4.3)/(sqrt(2a_{0)+1)}x+(43a_{0}+43)/(10sqrt(2a_{0)+1)}

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tangent of y=sqrt(2x+1)(4.3)

tangent of

y = 2 x + 1 (4.3)

y = \frac{4.3}{2 a _{0} + 1} x + \frac{43 a _{0} + 43}{10 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 \cdot 4.3)

Find the slope of

y = 2 x + 1 4.3 : \frac{d y}{d x} = \frac{4.3}{2 x + 1}

Compute the slope of

y = 2 x + 1 4.3

x = a_{0} : m = \frac{4.3}{2 a _{0} + 1}

Find the line with slope m=

\frac{4.3}{2 a _{0} + 1}

and passing through

(a_{0}, 2 a_{0} + 1 4.3) : y = \frac{4.3}{2 a _{0} + 1} x + \frac{43 a _{0} + 43}{10 2 a _{0} + 1}

y = \frac{4.3}{2 a _{0} + 1} x + \frac{43 a _{0} + 43}{10 2 a _{0} + 1}

\int x (x - 1)^{5} d x

\frac{\partial}{\partial x} (y^{\frac{1}{3}} x^{\frac{2}{3}})

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

f (x) = x^{3} - 2 x^{2} - 4 x - 6

\int (2 x + 5) e^{2 x + 5} d x

What is the tangent of y=sqrt(2x+1)(4.3) ?
The tangent of y=sqrt(2x+1)(4.3) is y=(4.3)/(sqrt(2a_{0)+1)}x+(43a_{0}+43)/(10sqrt(2a_{0)+1)}