What is the tangent of 1/(sqrt(x)),\at x=9 ?

The tangent of 1/(sqrt(x)),\at x=9 is y=-1/54 x+1/2

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tangent of 1/(sqrt(x)),\at x=9

tangent of

\frac{1}{x}, at x = 9

y = - \frac{1}{54} x + \frac{1}{2}

Solution steps

Find the tangent point:

(9, \frac{1}{3})

Find the slope of

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

Compute the slope of

y = \frac{1}{x}

x = 9 : m = - \frac{1}{54}

Find the line with slope m=

- \frac{1}{54}

and passing through

(9, \frac{1}{3}) : y = - \frac{1}{54} x + \frac{1}{2}

y = - \frac{1}{54} x + \frac{1}{2}

\frac{d}{d x} (x (x - 1))

x \to 0 + lim (tan^{x} (5 x))

x y^{^{'}} + (1 - x) y = 1

\int e^{\frac{x ^{2}}{2}} x d x

x \to \infty lim (x \cdot e^{- \frac{x}{2}})

What is the tangent of 1/(sqrt(x)),\at x=9 ?
The tangent of 1/(sqrt(x)),\at x=9 is y=-1/54 x+1/2