# Find the equation of the tangent at the given point

1. Jun 20, 2010

1. The problem statement, all variables and given/known data
For each function, find the equation of the tangent at the given point

a) f(x)= 1/ squareroot(2x+1)
at x=4

3. The attempt at a solution
I'm fairly lost, I understand I'm looking for an equation of a tangent, but I don't understand where to begin, find the derivative? Do I find the y value, then sub it back into y=mx+b?

2. Jun 20, 2010

### tiny-tim

(have a square-root: √ )

Yes, first use the derivative to find m, and then use the x and y at the given point to find b.

3. Jun 20, 2010

### malicx

you need two pieces of information: The slope of the tangent line and the y-intercept. The y-intercept should be easy enough. What do you know about the derivative of a function and the original graph?

4. Jun 20, 2010

First, yes I did try the derivative (which was y'= -x/ (√ 2x+1)^2 , if someone can verify that'd be great I used the quotient rule) , do I sub in my x value into that derivative? Still though, I wouldn't have my y value.

I've given you all the information the question gives. Sorry I'm confused about it too, which is why I used this forum. As I said in my other post, I found the derivative of the original function.

5. Jun 20, 2010

### tiny-tim

No, you only use the derivative to get m.

Once you have m, you should then put the x and y values (and m) into the original equation.
hmm … I've no idea how you got that.

Try using the chain rule instead (with u = 2x+1).

6. Jun 21, 2010

So with the chain rule I should get u= 2x+1, and then to get my y value sub x into the original function? Sorry if it seems like I don't get this at all, I really don't much.

7. Jun 21, 2010

### Staff: Mentor

No, you shouldn't get u = 2x + 1; tiny-tim was just reminding you that you could use the chain rule to get the derivative of y = 1/sqrt(2x + 1).
Here's what you need to do:
Find f'(x). As already pointed out, your first attempt was incorrect.
Evaluate f'(x) at x = 4. This will give you the slope m of the tangent line at the point (4, f(4)).

If you know the slope m of a line and a point (x0, y0) on the line, you can get the equation of the line by using the point-slope form of the line equation. It looks like this:
y - y0 = m(x - x0)

8. Jun 21, 2010

### tiny-tim

(just got up :zzz: …)
ok … try these examples …

you obviously know how to differentiate x5

how about differentiating (x + 3)5 ?

and differentiating (2x + 3)5 ?

if you can do that, just use the same method for (2x+1)-1/2

if you can't, tell us and we'll explain it.

9. Jun 21, 2010

### lanedance

i think you can probably use the chain & power laws, which can easily be derived from first principles (if need be), assume for some n>0
$$g(x) = \frac{1}{f(x)^n} = f(x)^{-n}$$

then the derivative is
$$g'(x) = \frac{dg(x)}{dx} =\frac{d}{dx}f(x)^{-n} = n f(x)^{(-n-1)} \frac{df(x)}{dx} = n f(x)^{-(n+1)}f'(x) = \frac{nf'(x)}{f(x)^{n+1}}$$

Last edited: Jun 21, 2010
10. Jun 21, 2010

### lanedance

when you know a line passes through x=a, y=b and has gradient m, the equation of the line is
$$y(x) = m(x-a)+ b$$

11. Jun 21, 2010

### Susanne217

Why make it more complex than it needs to be?

12. Jun 21, 2010

### lanedance

exactly - do you need a limit evaluation to find the derivative of a simple function...? i guess its all about perspective, or what people are comfortable with/find easiest

13. Jun 21, 2010

### Susanne217

Because my 1st semester College professor taught me if not asked to show the derivative then always use the definition of the derivative. Its easier and you don't risk getting an error then dealing with functions like this one! Specially if you are a newbee!

Last edited: Jun 21, 2010
14. Jun 21, 2010

### Staff: Mentor

Why do you make it more complex than it needs to be? Unless the problem specifically says to use the definition of the derivative, other methods are usually simpler.

Also, it is against the rules in this forum to provide complete solutions.

15. Jun 21, 2010

### Susanne217

Sorry...

16. Jun 21, 2010