1. ## Slope Tangent line

Hey guy wondering if someone could help me out here.

I need to find a formula for the slope of the tangent line to the graph of f at a general point x=xknot

and then use the formula obtained in first part to find the slope of the tangent line for the given value of xknot... heres the function f(x)= $\frac{1}{\sqrt{x}}$ ; xknot=4

I'm having trouble with the initial setup...

2. Originally Posted by nate19
Hey guy wondering if someone could help me out here.

I need to find a formula for the slope of the tangent line to the graph of f at a general point x=xknot

and then use the formula obtained in first part to find the slope of the tangent line for the given value of xknot... heres the function f(x)= $\frac{1}{\sqrt{x}}$ ; xknot=4

I'm having trouble with the initial setup...

In general, de slope of the tangent line to the graph of a function f at a general point $(x_0,y_0)$ is $f'(x_0)$ ,provided the function is derivable at that point.

Tonio

3. $\frac{d}{dx} \sqrt{x}^{-1}, x \neg 4= -\frac{1}{2x^{ \frac{3}{2}}}$

4. well thanks for the help guys but i'm not allowed to use derivatives or shortcuts yet.. the only formulas I can use are ..
lim x-->x0 f(x) - f(x0)/ x-x0

5. its just the same. the formula that u are given to use will actually give the derivative and thus the slope.

6. See attachment

7. finding derivatives ab-initio is cubersome