# On To Calculus - derivatives help

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• Jun 11th 2014, 05:04 PM
rhickman
On To Calculus - derivatives help
Two questions

Find the derivative of f(x)=$\displaystyle \sqrt{2x}$

I got the derivative $\displaystyle 1\frac2\sqrt2x$

Just confirming that's correct? I have to use the first principle rule to find it.

Second question is find the derivative of f(x) = $\displaystyle \frac{\4}{x+1}$

I believe the derivative should be $\displaystyle \frac{-\-4}{(x+1)^2}$

But when I work through the question (which I dont understand how to work through very well, because the example I was shown involved a 1 in the numerator not a 4)

I end up at $\displaystyle \frac{\1}{(x+1)^2}$

I'm not strong enough in latex to list these steps properly I dont think...

$\displaystyle \frac{\4}{x+h+1}-\frac{\4}{x+1}$

All of that, over h since the first principle is $\displaystyle \frac{f(x+h)-f(x)}{h}$

I dont know how to simplify any farther correctly (or get a 3 tier fraction on LaTex), I've tried following along with my example I was shown but it doesn't work correctly I dont think.

Any input is greatly appreciated, hopefully the LaTex helps to make my equations and stuff more clear.
• Jun 11th 2014, 05:41 PM
dokrbb
Re: On To Calculus - derivatives help
Quote:

Originally Posted by rhickman
Two questions

Find the derivative of f(x)=$\displaystyle \sqrt{2x}$

I got the derivative $\displaystyle 1\frac2\sqrt2x$

Just confirming that's correct? I have to use the first principle rule to find it.

Second question is find the derivative of f(x) = $\displaystyle \frac{\4}{x+1}$

I believe the derivative should be $\displaystyle \frac{\-4}{(x+1)^2}$

But when I work through the question (which I dont understand how to work through very well, because the example I was shown involved a 1 in the numerator not a 4)

I end up at $\displaystyle \frac{\1}{(x+1)^2}$

I'm not strong enough in latex to list these steps properly I dont think...

[tex]\frac{\4}{x+h+1}-\frac{\4}{x+1}[/t ex]

All of that, over h since the first principle is $\displaystyle \frac{f(x+h)-f(x)}{h}$

Hi there, in the first the derivative of $\displaystyle f(x)= \sqrt{2x}$ is actually $\displaystyle \frac{1}{\sqrt{2x}}$, do you see why?

and with the second, while applying $\displaystyle f'(x) = \frac{f(x)}{g(x)} = \frac{(f'(x)g(x))-(f(x)g'(x))}{(g(x))^2}$ you miss the negative sign,

cheers
• Jun 11th 2014, 06:49 PM
JeffM
Re: On To Calculus - derivatives help
Quote:

Originally Posted by rhickman
Two questions

Find the derivative of f(x)=$\displaystyle \sqrt{2x}$

I got the derivative $\displaystyle 1\frac2\sqrt2x$

Just confirming that's correct? I have to use the first principle rule to find it.

Second question is find the derivative of f(x) = $\displaystyle \frac{\4}{x+1}$

I believe the derivative should be $\displaystyle \frac{-\-4}{(x+1)^2}$

But when I work through the question (which I dont understand how to work through very well, because the example I was shown involved a 1 in the numerator not a 4)

I end up at $\displaystyle \frac{\1}{(x+1)^2}$

I'm not strong enough in latex to list these steps properly I dont think...

$\displaystyle \frac{\4}{x+h+1}-\frac{\4}{x+1}$

All of that, over h since the first principle is $\displaystyle \frac{f(x+h)-f(x)}{h}$

I dont know how to simplify any farther correctly (or get a 3 tier fraction on LaTex), I've tried following along with my example I was shown but it doesn't work correctly I dont think.

Any input is greatly appreciated, hopefully the LaTex helps to make my equations and stuff more clear.

It's hard to answer your questions because you do not tell us what general principles of differentiation you know.

Anyway, starting from first principles

$f(x) = \dfrac{4}{x + 1} \implies f(x + h) = \dfrac{4}{x + h + 1} \implies f(x + h) - f(x) = \dfrac{4}{x + h + 1} - \dfrac{4}{x + 1} =$

$\dfrac{4(x + 1)}{(x + 1)(x + h + 1)} - \dfrac{4(x + h + 1)}{(x + 1)(x + h + 1)} =\dfrac{4x + 4 - 4x - 4h - 4}{(x + 1)(x + h + 1)} = \dfrac{- 4h}{(x + 1)(x + h + 1)}.$

With me to here?

$\therefore \dfrac{f(x + h) - f(x)}{h} = \dfrac{1}{h} * \dfrac{- 4h}{(x + 1)(x + h + 1)} = \dfrac{- 4}{(x + 1)(x + h + 1)}.$

$Now\ \displaystyle \lim_{h \rightarrow 0}(- 4) = - 4\ and\ \lim_{h \rightarrow 0}(x + 1)(x + h + 1) = (x + 1)(x + 1) = (x + 1)^2.$

$So\ \displaystyle \lim_{h \rightarrow 0}\dfrac{- 4}{(x + 1)(x + h + 1)} = \dfrac{\displaystyle \lim_{h \rightarrow 0}(- 4)}{\displaystyle \lim_{h \rightarrow 0}(x + 1)(x + h + 1)} = \dfrac{- 4}{(x + 1)^2}.$

There are of course simpler ways to do this.
• Jun 11th 2014, 07:00 PM
rhickman
Re: On To Calculus - derivatives help
thank you very much Jeff, basically you started with the inverse so $\displaystyle \frac{-4h}{(x+1)(x+h+1)}$?

After sitting down for about an hour I got to the correct answer, still trying to figure out your steps but it will click after a few glances.

As for question one, I didn't do the coding correctly. I meant for question ones derivative to be $\displaystyle \frac{1}{2\sqrt2x}$ which is the derivative of $\displaystyle f(x)=\sqrt{2x}$
• Jun 11th 2014, 07:33 PM
rhickman
Re: On To Calculus - derivatives help
For question 1 is it supposed to be $\displaystyle f(x)=2x^\frac{1}{2}$

Which then becomes $\displaystyle 2*0.5x^{0.5-1}$

Which is then $\displaystyle 1x^{-0.5}$

....Im missing something I think, cause that gives me

$\displaystyle \frac{1}{\sqrt(x)}$
• Jun 11th 2014, 07:33 PM
JeffM
Re: On To Calculus - derivatives help
Quote:

Originally Posted by rhickman
thank you very much Jeff, basically you started with the inverse so $\displaystyle \frac{-4h}{(x+1)(x+h+1)}$?

After sitting down for about an hour I got to the correct answer, still trying to figure out your steps but it will click after a few glances.

As for question one, I didn't do the coding correctly. I meant for question ones derivative to be $\displaystyle \frac{1}{2\sqrt2x}$ which is the derivative of $\displaystyle f(x)=\sqrt{2x}$

I don't understand what you mean by inverse in this context.

There are two ways to find a derivative: (1) apply certain laws of derivatives, which is what you normally do after the first week or so of class, or (2) apply the basic definition of a derivative. I did the latter because I don't know what laws you have learned.

Anyway, the definitional approach actually gives you a procedure for finding a derivative.

Definition $f'(x) = \displaystyle \lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h}.$

Procedure (from definition rather than laws)

Write down f(x).

Write down f(x + h)

Write down f(x + h) - f(x)

Simplify that difference algebraically.

Divide the simplified difference by h (or multiply by 1/h).

Find the limit of that quotient as h approaches 0 using the laws of limits.

That is exactly the procedure I used.
• Jun 11th 2014, 07:39 PM
JeffM
Re: On To Calculus - derivatives help
Quote:

Originally Posted by rhickman
Or is it supposed to be $\displaystyle f(x)=2x^0.5$

Which then becomes $\displaystyle 2*0.5x^0.5-1$

Quite frankly, I am now not exactly sure even what the function is. I think it is $f(x) = \sqrt{2x}$. Is that right?

Please clarify what the function is and what method you used to find the derivative.
• Jun 11th 2014, 07:43 PM
rhickman
Re: On To Calculus - derivatives help
Thanks Jeff, I actually do this stuff through correspondence, or else I would just be asking my teacher (not to be rude, because I really do appreciate the help) I am working on learning a lot of laws in this lesson. I see what you did now so thats good.

As for the second question, yes $\displaystyle f(x)=\sqrt{2x}$ is the equation
• Jun 11th 2014, 07:58 PM
Prove It
Re: On To Calculus - derivatives help
Quote:

Originally Posted by rhickman
Thanks Jeff, I actually do this stuff through correspondence, or else I would just be asking my teacher (not to be rude, because I really do appreciate the help) I am working on learning a lot of laws in this lesson. I see what you did now so thats good.

As for the second question, yes $\displaystyle f(x)=\sqrt{2x}$ is the equation

Notice that \displaystyle \begin{align*} \sqrt{2x} = \sqrt{2}\,\sqrt{x} = 2^{\frac{1}{2}}\,x^{\frac{1}{2}} \end{align*}, this is what you would set f(x) as.
• Jun 11th 2014, 08:11 PM
rhickman
Re: On To Calculus - derivatives help
Quote:

Originally Posted by Prove It
Notice that \displaystyle \begin{align*} \sqrt{2x} = \sqrt{2}\,\sqrt{x} = 2^{\frac{1}{2}}\,x^{\frac{1}{2}} \end{align*}, this is what you would set f(x) as.

I see that but what I dont see is where I would go from there.

I think my answer should be $\displaystyle \frac{1}{\sqrt{2x}}$ at this point, but I may not have followed along clearly enough. My page is such a mess I can't even follow my own writing :p

EDIT: I may have understood it!

$\displaystyle 2^{0.5}x^{0.5}$

$\displaystyle =2^{0.5-1}x^{0.5-1}$

$\displaystyle 2^{-0.5}x^{-0.5}=\frac{1}{\sqrt{2}\sqrt{x}}$

Which finally becomes $\displaystyle =\frac{1}{\sqrt{2x}}$
• Jun 11th 2014, 08:12 PM
Prove It
Re: On To Calculus - derivatives help
Have you been taught the rule for derivatives: \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left( k\,x^n \right) = k\,n\,x^{n-1} \end{align*}?
• Jun 11th 2014, 08:23 PM
rhickman
Re: On To Calculus - derivatives help
Yes I have. For example

derivative of

$\displaystyle 3x^{2}$

$\displaystyle =2*3x^{(2-1)}$

$\displaystyle =6x$
• Jun 11th 2014, 08:41 PM
Prove It
Re: On To Calculus - derivatives help
Yes, so in your case you just have that \displaystyle \begin{align*} k = 2^{\frac{1}{2}} \end{align*} and \displaystyle \begin{align*} n = \frac{1}{2} \end{align*}. Apply the rule...
• Jun 11th 2014, 08:57 PM
rhickman
Re: On To Calculus - derivatives help
When I apply that rule I end up at

$\displaystyle (2)(\frac{1}{2})x^{0.5-1}$

Which gets me $\displaystyle x^{-0.5} or \frac{1}{\sqrt{x}}$
• Jun 11th 2014, 09:08 PM
rhickman
Re: On To Calculus - derivatives help
The Power Rule

$\displaystyle (\frac{1}{2})(2x)^{.5-1}*2$

$\displaystyle =2x^{-0.5} or \frac{1}{\sqrt{2x}}$

Or is that wrong?
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