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

The tangent of y= 2/(sqrt(x)),(9, 2/3) is y=-1/27 x+1

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tangent of y= 2/(sqrt(x)),(9, 2/3)

tangent of

y = \frac{2}{x}, (9, \frac{2}{3})

y = - \frac{1}{27} x + 1

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{2}{x}

(9, \frac{2}{3}) : m = - \frac{1}{27}

Find the line with slope m=

- \frac{1}{27}

and passing through

(9, \frac{2}{3}) : y = - \frac{1}{27} x + 1

y = - \frac{1}{27} x + 1

\int \frac{cos ( x )}{( cos ^{2} ( x ) ) ^{\frac{3}{2}}} d x

\int \frac{3 x - 24}{x ^{3} - 4 x ^{2} + 4 x} d x

\int_{0}^{2} e^{4 x} d x

\int \frac{x + 5}{( x + 3 ) ( x - 2 ) ^{2}} d x

x \to - 4 lim (3 ln (9) (7 - 5 x))

What is the tangent of y= 2/(sqrt(x)),(9, 2/3) ?
The tangent of y= 2/(sqrt(x)),(9, 2/3) is y=-1/27 x+1