What is the tangent of f(x)= 5/(sqrt(x)),\at x=4 ?

The tangent of f(x)= 5/(sqrt(x)),\at x=4 is y=-5/16 x+15/4

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tangent of f(x)= 5/(sqrt(x)),\at x=4

tangent of

f (x) = \frac{5}{x}, at x = 4

y = - \frac{5}{16} x + \frac{15}{4}

Solution steps

Find the tangent point:

(4, \frac{5}{2})

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 = 4 : m = - \frac{5}{16}

Find the line with slope m=

- \frac{5}{16}

and passing through

(4, \frac{5}{2}) : y = - \frac{5}{16} x + \frac{15}{4}

y = - \frac{5}{16} x + \frac{15}{4}

(lo g_{10} (x))^{^{'}}

63 e^{6 t}

\frac{\partial}{\partial x} (x^{4} - x^{2} y^{3})

6 - 5 x^{3}

6 x^{2} y^{2} d x + 4 x^{3} y d y = 0

What is the tangent of f(x)= 5/(sqrt(x)),\at x=4 ?
The tangent of f(x)= 5/(sqrt(x)),\at x=4 is y=-5/16 x+15/4