- #1

- 29

- 0

f(x) = Sqrt(x)/5 at (4,(2/5))

- Thread starter Victor Frankenstein
- Start date

- #1

- 29

- 0

f(x) = Sqrt(x)/5 at (4,(2/5))

- #2

- 759

- 0

*Simply find the derivative of the curve at that point, and use point-slope to find the line.

[tex] f\left( x \right) = \frac{\sqrt x}{5} \Rightarrow f\,'\left( x \right) = \frac{1}{{10\sqrt x }} [/tex]

*To find the slope of the tangent line, simply calculate [itex] f\,' ( 4 ) [/itex]:

[tex] f\,'\left( 4 \right) = \frac{1}{{20}} [/tex]

*Next, just use point-slope to represent the tangent line, which I'll call [itex] y [/itex]:

[tex] \frac{{x - 4}}{{20}} = y - \frac{2}{5} [/tex]

And it's algebra from there

[tex] f\left( x \right) = \frac{\sqrt x}{5} \Rightarrow f\,'\left( x \right) = \frac{1}{{10\sqrt x }} [/tex]

*To find the slope of the tangent line, simply calculate [itex] f\,' ( 4 ) [/itex]:

[tex] f\,'\left( 4 \right) = \frac{1}{{20}} [/tex]

*Next, just use point-slope to represent the tangent line, which I'll call [itex] y [/itex]:

[tex] \frac{{x - 4}}{{20}} = y - \frac{2}{5} [/tex]

And it's algebra from there

Last edited:

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

Victor Frankenstein said:

f(x) = Sqrt(x)/5 at (4,(2/5))

Are you saying that you are required to use the basic formula for the derivative: [itex] lim_{h->0}\frac{f(a+h)-f(a)}{h}[/itex] rather than the more specific formula (derived from that) that bomba923 used?

If so try multiplying both numerator and denominator of

[tex]\frac{\sqrt{a+h}-\sqrt{a}}{h}[/tex]

by

[tex]\sqrt{a+h}+ \sqrt{a}[/tex].

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 13

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 31

- Views
- 3K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 5K

- Replies
- 2

- Views
- 5K