Find the equation of the tangent at the given point

[Aske]

Homework Statement

For each function, find the equation of the tangent at the given point

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

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?

 

[tiny-tim]

Hi Aske!

(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.

 

[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?

 

[Aske]

Hi Aske!

(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.

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.

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?

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.

 

[tiny-tim]

… do I sub in my x value into that derivative? Still though, I wouldn't have my y value.

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.

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) ,

hmm … I've no idea how you got that.

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

 

[Aske]

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). ​

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.

 

[Mark44]

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.

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)

 

[tiny-tim]

Hi Aske!

(just got up :zzz: …)

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.

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.

 

[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
$$g'(x) = \frac{dg(x)}{dx} =\frac{d}{dx}f(x)^{-n} <br /> = n f(x)^{(-n-1)} \frac{df(x)}{dx} = n f(x)^{-(n+1)}f'(x) = \frac{nf'(x)}{f(x)^{n+1}}$$

 

[lanedance]

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

 

[Susanne217]

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

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

Why make it more complex than it needs to be?

 

[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

 

[Susanne217]

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

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!

 

[Mark44]

You first task is it find the right derivative.. Then I'm not asked to show the derivative I simply use the generalized formula for Calculus

You know that $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

and to find the derivative at x = a

then $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

and you simply get $$f'(4) = \lim_{h \to 0} \left(\frac{1}{h \cdot \sqrt{2 \cdot 4 + 2 \cdot h +1}} - \frac{1}{h \cdot \sqrt{2 \cdot 4 +1}}}\right) = -\frac{1}{27}$$

where f'(4) is the slope of the tangent for f(x) at x = 4

and using an other know equation for the tangent

$$y = f(a) + f'(a) \cdot (x-a)$$

We get $$y = \frac{1}{3} - \frac{1}{27} \cdot (x-4)$$

cheers ;)

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.

 

[Susanne217]

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.

Sorry...

 

[Aske]

Its ok, I got to ask at school about this question, I was shown how the chain rule was used (I haven't used it much tbh), so I now understand this question. A mod can now close this thread if needed.