What is the tangent of y=sqrt(x+3),(6,3) ?

The tangent of y=sqrt(x+3),(6,3) is y= 1/6 x+2

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tangent of y=sqrt(x+3),(6,3)

tangent of

y = x + 3, (6, 3)

y = \frac{1}{6} x + 2

Solution steps

Find the slope of

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

Compute the slope of

y = x + 3

(6, 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(6, 3) : y = \frac{1}{6} x + 2

y = \frac{1}{6} x + 2

\frac{d y}{d x} = 2 e^{x - y}

\int \frac{x ^{3}}{x ^{4} - x ^{2} + 2} d x

f (x) = (3 x + 7) e^{x}

x^{2} + 10

\frac{d}{d x} (1.394 (1.006)^{x})

What is the tangent of y=sqrt(x+3),(6,3) ?
The tangent of y=sqrt(x+3),(6,3) is y= 1/6 x+2