What is the tangent of f(x)= 5/(sqrt(x)),\at x=1 ?

The tangent of f(x)= 5/(sqrt(x)),\at x=1 is y=-5/2 x+15/2

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tangent of f(x)= 5/(sqrt(x)),\at x=1

tangent of

f (x) = \frac{5}{x}, at x = 1

y = - \frac{5}{2} x + \frac{15}{2}

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = \frac{5}{x} : \frac{df ( x )}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{5}{x}

x = 1 : m = - \frac{5}{2}

Find the line with slope m=

- \frac{5}{2}

and passing through

(1, 5) : y = - \frac{5}{2} x + \frac{15}{2}

y = - \frac{5}{2} x + \frac{15}{2}

\frac{1}{2 s + 3}

\frac{\partial}{\partial y} (x y e^{2 y})

x \to 1 lim (arcsec (x))

\int cos (3 t) + sin (4 t) d t

\frac{d}{d x} (x^{3} + \frac{1}{x})

What is the tangent of f(x)= 5/(sqrt(x)),\at x=1 ?
The tangent of f(x)= 5/(sqrt(x)),\at x=1 is y=-5/2 x+15/2