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

The tangent of f(x)= 1/(sqrt(x)),\at x=a is y=-1/(2a^{3/2)}x+3/(2sqrt(a))

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

tangent of

f (x) = \frac{1}{x}, at x = a

y = - \frac{1}{2 a ^{\frac{3}{2}}} x + \frac{3}{2 a}

Solution steps

Find the tangent point:

(a, \frac{1}{a})

Find the slope of

f (x) = \frac{1}{x} : \frac{df ( x )}{d x} = - \frac{1}{2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{x}

x = a : m = - \frac{1}{2 a ^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{2 a ^{\frac{3}{2}}}

and passing through

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

y = - \frac{1}{2 a ^{\frac{3}{2}}} x + \frac{3}{2 a}

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

\int \frac{e ^{x}}{1 + e ^{x} + e ^{2 x}} d x

\int 6 x - 7 d x

\frac{\partial}{\partial x} (- e^{- o} \cdot cos (o))

\frac{\partial}{\partial y} (ln (x - 3 y))

What is the tangent of f(x)= 1/(sqrt(x)),\at x=a ?
The tangent of f(x)= 1/(sqrt(x)),\at x=a is y=-1/(2a^{3/2)}x+3/(2sqrt(a))