What is the tangent of f(x)= 1/(sqrt(x)),\at x=25 ?

The tangent of f(x)= 1/(sqrt(x)),\at x=25 is y=-1/250 x+3/10

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tangent of f(x)= 1/(sqrt(x)),\at x=25

tangent of

f (x) = \frac{1}{x}, at x = 25

y = - \frac{1}{250} x + \frac{3}{10}

Solution steps

Find the tangent point:

(25, \frac{1}{5})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{250}

Find the line with slope m=

- \frac{1}{250}

and passing through

(25, \frac{1}{5}) : y = - \frac{1}{250} x + \frac{3}{10}

y = - \frac{1}{250} x + \frac{3}{10}

f (x) = (4 x - 3)^{\frac{1}{2}}

y^{^{''}} + y = - sin (2 x), y (0) = 0

\int \frac{1}{11 x - 5} d x

4 y^{^{''}} + 28 y = 0

\frac{\partial}{\partial x} (f (x, y) x (x, y))

What is the tangent of f(x)= 1/(sqrt(x)),\at x=25 ?
The tangent of f(x)= 1/(sqrt(x)),\at x=25 is y=-1/250 x+3/10