What is the tangent of f(x)= 3/(sqrt(x)),\at x=a ?

The tangent of f(x)= 3/(sqrt(x)),\at x=a is y=-3/(2a^{3/2)}x+9/(2sqrt(a))

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tangent of f(x)= 3/(sqrt(x)),\at x=a

tangent of

f (x) = \frac{3}{x}, at x = a

y = - \frac{3}{2 a ^{\frac{3}{2}}} x + \frac{9}{2 a}

Solution steps

Find the tangent point:

(a, \frac{3}{a})

Find the slope of

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

Compute the slope of

f (x) = \frac{3}{x}

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

Find the line with slope m=

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

and passing through

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

y = - \frac{3}{2 a ^{\frac{3}{2}}} x + \frac{9}{2 a}

\frac{d}{d x} (sin (x) x^{2})

\frac{d}{d x} (e^{x} - 2 cos (x))

\frac{\partial}{\partial x} (4 x^{- \frac{1}{2}} - \frac{6}{x})

\frac{\partial}{\partial x} (ln (9 + x^{2} + 3 y^{2}))

\frac{d}{d x} (\frac{6}{( x - 1 ) ^{4}})

What is the tangent of f(x)= 3/(sqrt(x)),\at x=a ?
The tangent of f(x)= 3/(sqrt(x)),\at x=a is y=-3/(2a^{3/2)}x+9/(2sqrt(a))