# Differentiation Question

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• May 22nd 2008, 05:04 PM
Pikeman85
Differentiation Question
So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

however this is obviously not a correct calculation - what was I doing wrong?
• May 22nd 2008, 05:12 PM
r_maths
This site helped me alot.

http://www.mathhelpforum.com/math-he...step-step.html
solving derivatives step-by-step

Just type in sqrt[(5x+1)/(4x^2+1)].
• May 22nd 2008, 05:15 PM
Mathstud28
Quote:

Originally Posted by Pikeman85
So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

however this is obviously not a correct calculation - what was I doing wrong?

$y=\sqrt{\frac{5x+1}{4x^2+1}}\Rightarrow{y^2=\frac{ 5x+1}{4x^2+1}}$

Differentiating we get

$2y\cdot{y'}=\frac{-20x^2-8x+5}{(4x^2+1)^2}$

So remembering that $y=\sqrt{\frac{5x+1}{4x^2+1}}$

$y'=\frac{-20x^2-8x+5}{2\sqrt{\frac{5x+1}{4x^2+1}}(4x^2+1)^2}$

so $y'(2)=\frac{-91\sqrt{187}}{6350}\approx{-.195723}$
• May 22nd 2008, 05:20 PM
Chris L T521
Quote:

Originally Posted by Pikeman85
So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

however this is obviously not a correct calculation - what was I doing wrong?

$y=\sqrt{\frac{5x+1}{4x^2+1}}$

$y^{/}=\frac{1}{2}\left(\frac{5x+1}{4x^2+1}\right)^{-\frac{1}{2}}\cdot \left(\frac{(8x)(5x+1)-(4x^2+1)(5)}{(4x^2+1)^2}\right)$

Try taking it from here.
• May 22nd 2008, 05:22 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
$y=\sqrt{\frac{5x+1}{4x^2+1}}$

$y^{/}=\frac{1}{2}\left(\frac{5x+1}{4x^2+1}\right)^{-\frac{1}{2}}\cdot \left(\frac{(8x)(5x+1)-(4x^2+1)(5)}{(4x^2+1)^2}\right)$

Try taking it from here.

Why do you put

$y^{/}$ instead of $y'$

I think the latter looks nicer...not that it matters...I am just curious
• May 22nd 2008, 05:23 PM
martinr
RE:diff
I punched you question into matlab, the derivative of your function

f(x) = sqrt(((5*x+1)/(4*x^2+1)))

is

f '(x) = 1/2/((5*x+1)/(4*x^2+1))^(1/2)*(5/(4*x^2+1)-8*(5*x+1)/(4*x^2+1)^2*x)

Im certain its 100% correct since a computer calculated it, if I'm wrong, let me know(Rock)
• May 22nd 2008, 05:36 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
Why do you put

$y^{/}$ instead of $y'$

I think the latter looks nicer...not that it matters...I am just curious

Its just a habbit. I do this a lot when I use mathtype, for it looks better that (').
• May 22nd 2008, 05:42 PM
Chris L T521
Quote:

Originally Posted by martinr
I punched you question into matlab, the derivative of your function

f(x) = sqrt(((5*x+1)/(4*x^2+1)))

is

f '(x) = 1/2/((5*x+1)/(4*x^2+1))^(1/2)*(5/(4*x^2+1)-8*(5*x+1)/(4*x^2+1)^2*x)

Im certain its 100% correct since a computer calculated it, if I'm wrong, let me know(Rock)

Mathematica gave me this:

I assume its the same??
• May 22nd 2008, 06:01 PM
Pikeman85
Hrm, here is another problem, this one is VERY difficult for me to solve.

[IMG]file:///D:/DOCUME%7E1/ANDREW%7E2.MAI/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=11, f(a)=3,f'(a)=10, f'(a^{5})=5.

Find F'(a)= and G'(a)=

• May 22nd 2008, 06:25 PM
Mathstud28
Quote:

Originally Posted by Pikeman85
Hrm, here is another problem, this one is VERY difficult for me to solve.

[IMG]file:///D:/DOCUME%7E1/ANDREW%7E2.MAI/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=11, f(a)=3,f'(a)=10, f'(a^{5})=5.

Find F'(a)= and G'(a)=

$F(x)=f(x^5)$

differentiating using the chain rule we get

$F'(x)=f'(x^5)\cdot{5x^4}$

so $F'(a)=f'(a^5)\cdot{5a^4}$

and since they give us $f(a^5)=5$ and $a^4=11$

we have $F'(5)=5\cdot{5}\cdot{11}=275$
• May 22nd 2008, 07:11 PM
Pikeman85
I'd generally prefer it if people gave me more hints than answers. Oh work through part of a problem maybe, but eh. I want to figure this stuff out myself so that I actually KNOW it.

Now this one - this one is difficult for me (and thanks for everyone that has helped me this time and the time before! I'm actually getting some of the differentiation stuff quite a bit better now - still need to work on the Chain Rule, and a couple other things, but I have many of the others)

Ok so here's the problem I have now:

f(x) =
{ x^2 + 6x + 13 X (less than or equal to sign) 2
{ ax + b X (greater than) 2

Find a and b such that the function is differentiable everywhere. So I'm looking for continuity, and I know that at 2, the top equation = 28.

How do I go about solving this?
• May 22nd 2008, 07:17 PM
Mathstud28
Quote:

Originally Posted by Pikeman85
I'd generally prefer it if people gave me more hints than answers. Oh work through part of a problem maybe, but eh. I want to figure this stuff out myself so that I actually KNOW it.

Now this one - this one is difficult for me (and thanks for everyone that has helped me this time and the time before! I'm actually getting some of the differentiation stuff quite a bit better now - still need to work on the Chain Rule, and a couple other things, but I have many of the others)

Ok so here's the problem I have now:

f(x) =
{ x^2 + 6x + 13 X (less than or equal to sign) 2
{ ax + b X (greater than) 2

Find a and b such that the function is differentiable everywhere. So I'm looking for continuity, and I know that at 2, the top equation = 28.

How do I go about solving this?

You dont have to CAPITALIZE things (Surprised)...I understand

Ok so you want a hint but not an answer

if f(x) is differentiable at c then

$\lim_{x\to{c^{+}}}f(x)=\lim_{x\to{c^{-}}}f(x)$

and $\lim_{x\to{c^{-}}}f'(x)=\lim_{x\to{c^{+}}}f'(x)$
• May 22nd 2008, 07:27 PM
Pikeman85
Yes, I know that the two points have to touch, the problem is I am not sure how I am supposed to calculate which two points do touch :(
• May 22nd 2008, 07:29 PM
Mathstud28
Quote:

Originally Posted by Pikeman85
Yes, I know that the two points have to touch, the problem is I am not sure how I am supposed to calculate which two points do touch :(

Ok, so what you must do is this for it to be differentiable at the problem spot x=2

f(x) to the left of two must equal f(x) from the right of two

and f'(x) from the left of two must equal f'(x) from the right of two

so now you have to equations with two unknowns. Solve for the unknowns
• May 22nd 2008, 07:43 PM
Pikeman85
Still not getting it. I'm very dense about a lot of the simpler bits. I've got most of the differentiation ones done, but this one eludes me :(
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