What is the tangent of y=sqrt(x),(9,3) ?

The tangent of y=sqrt(x),(9,3) is y= 1/6 x+3/2

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tangent of y=sqrt(x),(9,3)

tangent of

y = x, (9, 3)

y = \frac{1}{6} x + \frac{3}{2}

Solution steps

Find the slope of

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

Compute the slope of

y = x

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

Find the line with slope m=

\frac{1}{6}

and passing through

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

y = \frac{1}{6} x + \frac{3}{2}

n = 6 \sum \infty \frac{5 ^{n}}{1 2 ^{n}}

3 y^{^{''}} + 17 y^{^{'}} - 6 y = 0

\frac{d}{d x} (\frac{1}{x ^{4}} + \frac{1}{x ^{2}} - x^{3})

\int (x^{2} - 5)^{5} \cdot 2 x d x

y^{^{''}} = x e^{x}

What is the tangent of y=sqrt(x),(9,3) ?
The tangent of y=sqrt(x),(9,3) is y= 1/6 x+3/2