What is the tangent of y=sqrt(x),(81,9) ?

The tangent of y=sqrt(x),(81,9) is y= 1/18 x+9/2

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tangent of y=sqrt(x),(81,9)

tangent of

y = x, (81, 9)

y = \frac{1}{18} x + \frac{9}{2}

Solution steps

Find the slope of

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

Compute the slope of

y = x

(81, 9) : m = \frac{1}{18}

Find the line with slope m=

\frac{1}{18}

and passing through

(81, 9) : y = \frac{1}{18} x + \frac{9}{2}

y = \frac{1}{18} x + \frac{9}{2}

x \to 0 lim (sin (x))

\int \frac{1}{x y} d x

\frac{\partial}{\partial r} (e^{r} cos (t))

x^{2} (\frac{d y}{d x}) - x y + 3 y^{2} = 0

\int_{0}^{2} x^{3} x^{2} + 4 d x

What is the tangent of y=sqrt(x),(81,9) ?
The tangent of y=sqrt(x),(81,9) is y= 1/18 x+9/2