Finding derivatives of functions

[TommG]

Need to find derivative

f(s) = [(√s) -1]/[(√s) + 1]

answer in book is f'(s) = 1/[√s(√s+1)^2]

 

[SteamKing]

Your attempt at finding the derivative apparently consists of re-writing f(s) and replacing the square root signs with exponents.

Do you know how to take the derivative of this function: f(s) = s^(1/2)?

 

[TommG]

Your attempt at finding the derivative apparently consists of re-writing f(s) and replacing the square root signs with exponents.

Do you know how to take the derivative of this function: f(s) = s^(1/2)?

yes I do. I deleted my attempt because I don't think I can write it well enough so people can understand it. Is it ok if I post a pic of it?

 

[Ray Vickson]

yes I do. I deleted my attempt because I don't think I can write it well enough so people can understand it. Is it ok if I post a pic of it?

You should post it. If you use plain text you need to avoid ambiguity by use of parentheses; for example, a/b*c can mean either (a/b)*c or a/(b*c), so using ( ) is important. If you have several levels of nested parentheses, you can use different styles, such as [a + b*(c+d)]/(e + f*g) instead of (a + b*(c+d))/(e+f*g) if you think it looks clearer.

However, a much better way is to use LaTeX. Use [i tex ] xxxx [/i tex ] (remove spaces) to put the material xxx in line, such as $a = \pm \sqrt{b}$. Alternatively, you can use # # xxx # # (remove spaces between the two #s); for example, $a \geq b+c^2$. If you want displayed equations instead, use [ t ex ] xxxx [/ t e x] (no spaces) or $ $ xxxx $ $ (no spaces between $s); for example, here are two displayed expressions entered inside a single [t ex ] ... [/t ex ] pair:
$$a = \pm \sqrt{b}\\ \alpha \geq \frac{\beta + \gamma^2}{\beta - \gamma^2}$$

As for the commands themselves, refer to the LaTeX tutorial that you can find under the forum 'Site Info' -> 'Frequently Asked Questions' -> '2. How to type mathematical equations'. By far the easiest way is to examine a posting that has nice equations in it and right click on the equation (or go to the menu item display math as latex commands) and just look at what people typed.

 

[TommG]

Well I figured out this problem but need help with another.
Thanks ray for the advice.

I could use help with a new problem.


Have to find the derivative.

$v = \frac {1 + x - 4\sqrt{x}} x$

My attempt

$v = \frac {1 + x - 4\sqrt{x}} x$

$v = \frac {x(1-\frac{2}{√x}) - (1+x -4 √x)}{x^2}$

$v = \frac {x-\frac{2x}{√x} -1-x +4 √x}{x^2}$

$v = \frac {-\frac{2x}{√x} -1- +4 √x}{x^2}$ don't know where to go from here


answer in book is $$v' = \frac {-1}{x^2}+ 2x ^{-3/2}$$

 

[Ray Vickson]

Well I figured out this problem but need help with another.
Thanks ray for the advice.

I could use help with a new problem.


Have to find the derivative.

$v = \frac {1 + x - 4\sqrt{x}} x$

My attempt

$v = \frac {1 + x - 4\sqrt{x}} {x}$

$v = \frac {x(1-\frac{2}{√x}) - (1+x -4 √x)}{x^2}$

$v = \frac {x-\frac{2x}{√x} -1-x +4 √x}{x^2}$

$v = \frac {-\frac{2x}{√x} -1- +4 √x}{x^2}$ don't know where to go from here


answer in book is $$v' = \frac {-1}{x^2}+ 2x ^{-3/2}$$

You are doing it the hard way (and I have not checked your work). It is much easier to expand out $v$ and so write
$$v = \frac {1 + x - 4\sqrt{x}}{ x} = \frac{1}{x} + 1 - \frac{4}{\sqrt{x}} = x^{-1} + 1 - 4 x^{-1/2}$$
and then differentiate each term separately. If you want, you can re-combine the results in the end.

BTW: in "\ frac{ } " you forgot the second { } (so you had "\ frac {y} x " instead of "\ frac {y}{x}". I am surprised it worked at all, but it does, as shown by the following test $\frac{y}x$. Nevertheless, you should NEVER forget that final {} because if you have something more than one letter it won't work: "\ frac{y}x\sin(x) " produces $\frac{y}x\sin(x)$ instead of the wanted $\frac{y}{x \sin(x)}$.

 

[TommG]

You are doing it the hard way (and I have not checked your work). It is much easier to expand out $v$ and so write
$$v = \frac {1 + x - 4\sqrt{x}}{ x} = \frac{1}{x} + 1 - \frac{4}{\sqrt{x}} = x^{-1} + 1 - 4 x^{-1/2}$$
and then differentiate each term separately. If you want, you can re-combine the results in the end.

BTW: in "\ frac{ } " you forgot the second { } (so you had "\ frac {y} x " instead of "\ frac {y}{x}". I am surprised it worked at all, but it does, as shown by the following test $\frac{y}x$. Nevertheless, you should NEVER forget that final {} because if you have something more than one letter it won't work: "\ frac{y}x\sin(x) " produces $\frac{y}x\sin(x)$ instead of the wanted $\frac{y}{x \sin(x)}$.

I have to do it the hard way.

I have to follow the derivative quotient rule.

Still need help if anyone could help out.

 

[jbunniii]

BTW: in "\ frac{ } " you forgot the second { } (so you had "\ frac {y} x " instead of "\ frac {y}{x}". I am surprised it worked at all, but it does, as shown by the following test $\frac{y}x$. Nevertheless, you should NEVER forget that final {} because if you have something more than one letter it won't work: "\ frac{y}x\sin(x) " produces $\frac{y}x\sin(x)$ instead of the wanted $\frac{y}{x \sin(x)}$.

Interestingly, if you have just one letter in the numerator and in the denominator, you don't need braces at all:

##\frac a b##

is displayed as $\frac a b$. I never realized that would work. I guess it makes sense in retrospect.

 

[Ray Vickson]

I have to do it the hard way.

I have to follow the derivative quotient rule.

Still need help if anyone could help out.

(1) Why must you do it the hard way? Where in the original question does it tell you that?
(2) Why did you put everything over a common denominator? The quotient rule says that
$$\left( \frac{u}{v} \right)' = \frac{u'}{v} - \frac{u v'}{v^2}$$
but you wrote
$$\left( \frac{u}{v} \right)' = \frac{u' v - u v'}{v^2}$$
for no good reason that I can see. It is correct but unnecessary.
(3) You should combine and simplify your previous final answer.
(4) The book's answer does NOT involve a common denominator, so you need to 'undo' your final answer, or else re-write the book's answer in a form similar to yours, so you can compare the two.

 

[TommG]

(1) Why must you do it the hard way? Where in the original question does it tell you that?
(2) Why did you put everything over a common denominator? The quotient rule says that
$$\left( \frac{u}{v} \right)' = \frac{u'}{v} - \frac{u v'}{v^2}$$
but you wrote
$$\left( \frac{u}{v} \right)' = \frac{u' v - u v'}{v^2}$$
for no good reason that I can see. It is correct but unnecessary.
(3) You should combine and simplify your previous final answer.
(4) The book's answer does NOT involve a common denominator, so you need to 'undo' your final answer, or else re-write the book's answer in a form similar to yours, so you can compare the two.

I have to do it the hard because my teacher said I have to when he gives us a quiz.

I put everything over a common denominator because of the rule in the book.
Here is a pic I took from my book.
https://www.dropbox.com/s/ca3l50hmgh56n41/IMG_20140606_121016.jpg

 

[HallsofIvy]

Well I figured out this problem but need help with another.
Thanks ray for the advice.

I could use help with a new problem.


Have to find the derivative.

$v = \frac {1 + x - 4\sqrt{x}} x$

My attempt

$v = \frac {1 + x - 4\sqrt{x}} x$

$v = \frac {x(1-\frac{2}{√x}) - (1+x -4 √x)}{x^2}$

$v = \frac {x-\frac{2x}{√x} -1-x +4 √x}{x^2}$

$v = \frac {-\frac{2x}{√x} -1- +4 √x}{x^2}$ don't know where to go from here

Surely you know that $\frac{x}{\sqrt{x}}= \sqrt{x}$?
What you have is the same as
$$\frac{-2\sqrt{x}- 1+ 4\sqrt{x}}{x^2}= \frac{2\sqrt{x}- 1}{x^2}= 2\frac{\sqrt{x}}{x^2}- \frac{1}{x^2}$$
And, of course, $\sqrt{x}/x^2= x^{1/2- 2}$

answer in book is $$v' = \frac {-1}{x^2}+ 2x ^{-3/2}$$

 

[Fredrik]

The quotient rule says that
$$\left( \frac{u}{v} \right)' = \frac{u'}{v} - \frac{u v'}{v^2}$$
but you wrote
$$\left( \frac{u}{v} \right)' = \frac{u' v - u v'}{v^2}$$

The second equation is the quotient rule as it was taught to me.

 

[SammyS]

The second equation is the quotient rule as it was taught to me.

Yup! (I agree with Fredrik -- if anyone cares.)

low*D(high) minus high*D(low) all over low squared.