What is the tangent of y=sqrt(2x+1),(4,3) ?

The tangent of y=sqrt(2x+1),(4,3) is y= 1/3 x+5/3

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tangent of y=sqrt(2x+1),(4,3)

tangent of

y = 2 x + 1, (4, 3)

y = \frac{1}{3} x + \frac{5}{3}

Solution steps

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}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(4, 3) : y = \frac{1}{3} x + \frac{5}{3}

y = \frac{1}{3} x + \frac{5}{3}

2 t^{- \frac{1}{4}}

\int \frac{x ^{2}}{e ^{\frac{x}{3}}} d x

5 \cdot t

\int \frac{1}{x ^{2} x ^{2} + 25} d x

\frac{\partial}{\partial y} (\frac{x}{y})

What is the tangent of y=sqrt(2x+1),(4,3) ?
The tangent of y=sqrt(2x+1),(4,3) is y= 1/3 x+5/3