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

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

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

tangent of

y = \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

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 = 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{\partial}{\partial x} (4 tan (x y))

f (x) = tan (2 x)

f (x) = x x, at x = 36

\int \frac{sec ( x ) + cos ( x )}{5 cos ( x )} d x

\frac{1}{1 + 2 x}, 0

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