What is the tangent of y=6sqrt(x),(9,18) ?

The tangent of y=6sqrt(x),(9,18) is y=x+9

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tangent of y=6sqrt(x),(9,18)

tangent of

y = 6 x, (9, 18)

y = x + 9

Solution steps

Find the slope of

y = 6 x : \frac{d y}{d x} = \frac{3}{x}

Compute the slope of

y = 6 x

(9, 18) : m = 1

Find the line with slope m=

1

and passing through

(9, 18) : y = x + 9

y = x + 9

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

x \to 0 lim (\frac{sin ( x )}{2 x})

\frac{d}{d x} (2 cos (x) + 2 x sin (x))

\frac{d y}{d x} + y = 2 x^{2}

\int (4 x + 5) d x

What is the tangent of y=6sqrt(x),(9,18) ?
The tangent of y=6sqrt(x),(9,18) is y=x+9