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

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

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

tangent of

y = x + 2, (- 1, 1)

y = \frac{1}{2} x + \frac{3}{2}

Solution steps

Find the slope of

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

Compute the slope of

y = x + 2

(- 1, 1) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(- 1, 1) : y = \frac{1}{2} x + \frac{3}{2}

y = \frac{1}{2} x + \frac{3}{2}

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

x \to - 4 + lim (\frac{- 2}{x ^{2} - 16})

\frac{\partial}{\partial x} (ln (\frac{x}{4}))

\int \frac{32 x ^{2}}{( x - 12 ) ( x + 4 ) ^{2}} d x

\frac{x ^{4}}{5 - x ^{3}}

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