Implicit Differation and Chain Rule

• Jun 7th 2008, 11:44 AM
PensFan10
Implicit Differation and Chain Rule
First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
The problem I need help with is:

x^2 + xy - y^2 = 4 Solver for y'
I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).

Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

Thank you!!!
• Jun 7th 2008, 11:51 AM
Moo
Hello,

Quote:

Originally Posted by PensFan10
First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
The problem I need help with is:

x^2 + xy - y^2 = 4 Solver for y'
I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).

Your only mistake (that I found by analysing your answer, so sorry if you did it right..) is that the derivative of xy has to be taken with the product rule :

$(xy)'=(x)'y+x(y)'=y+xy'$

As soon as the variable intervenes in x, y, z, t, or whatever, you have to consider it as a function, not like a variable, so you're likely to use the product, quotient or chain rule =)

Quote:

Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

Thank you!!!
$\sqrt{\frac{2x}{x^2+1.8}}=\sqrt{u(x)}$

The chain rule says that the derivative is $\frac{u'(x)}{2 \sqrt{u(x)}}$

So calculate $u'(x)$ with the quotient rule.

Always do things step by step. It's pretty tricky for someone who is not used to differentiating to do it without taking u(x).

:)
• Jun 7th 2008, 11:59 AM
PensFan10
Thanks so much
That was stupid of me to mess up the xy (product rule). Doh! Thanks for catching that mistake.

Thanks a lot for the help!!!
• Jun 7th 2008, 12:11 PM
PensFan10
One more quick question:
So I solved chain rule problem out and got an answer of:

(1)/(XSqrt(2x/x^2+1.8))

Just curious if I did it correct. Once again, Thanks!
• Jun 7th 2008, 12:17 PM
Moo
Quote:

Originally Posted by PensFan10
So I solved chain rule problem out and got an answer of:

(1)/(XSqrt(2x/x^2+1.8))

Just curious if I did it correct. Once again, Thanks!

Hmm; that's a weird one.

What do you find as the derivative $u'(x)$ ?
Because I doubt it is $\frac 2x$ /:

Can you show your working so that I can tell you where the mistake(s) is(are) ? (Nod)
• Jun 7th 2008, 12:29 PM
PensFan10
Of course
Okay, I see how I messed up.
Heres my current solution attempt :P

For u'(x) I have this:

2x/(x^2+1.8)

Quotient rule says: gf'-fg'/g^2

When I plug my info in I got:

(x^2+1.8)(2)-(2x)(2x)/(x^2+1.8)^2
Simplified:
2x^2+3.6-4x^2/(x^2+1.8)^2

=-2x^2+3.6/(x^2+1.8)^2

Now I plug that into u'(x)/(2Sqrt u(x))

I got

-2x^2+3.6
-----------------
(x^2+1.8)^2
------------------
2Sqrt(2x/x^2+1.8)

Might be just me, but I think I did something wrong again.
• Jun 7th 2008, 12:35 PM
Moo
Quote:

Originally Posted by PensFan10
Okay, I see how I messed up.
Heres my current solution attempt :P

For u'(x) I have this:

2x/(x^2+1.8)

Quotient rule says: gf'-fg'/g^2

When I plug my info in I got:

(x^2+1.8)(2)-(2x)(2x)/(x^2+1.8)^2
Simplified:
2x^2+3.6-4x^2/(x^2+1.8)^2

=-2x^2+3.6/(x^2+1.8)^2

Now I plug that into u'(x)/(2Sqrt u(x))

I got

-2x^2+3.6
-----------------
(x^2+1.8)^2
------------------
2Sqrt(2x/x^2+1.8)

Might be just me, but I think I did something wrong again.

Ignoring the missing parentheses, this is correct :)

Note that you can write $-2x^2+3.6=2(-x^2+1.8)$ and simplify the fraction by 2 (Nod)

The solution is :

$\frac{2(-x^2+1.8)}{2(x^2+1.8)^2 \sqrt{\frac{2x}{x^2+1.8}}}=\frac{-x^2+1.8}{(x^2+1.8)^2 \cdot \sqrt{2x} \cdot (x^2+1.8)^{\color{red}-\frac 12}}$

-----------------------------
do you understand the red part ?

$\frac{1}{a^b}=a^{-b}$, and $\sqrt{a}=a^{\frac 12}$
-----------------------------

Simplifying the powers by this rule : $a^b a^c=a^{b+c}$, we get :

$f'(x)=\frac{-x^2+1.8}{\sqrt{2x} \cdot (x^2+1.8)^{\frac 32}}$

Now it depends on how you want to display the result :)
• Jun 7th 2008, 12:39 PM
PensFan10
Awesome!
Yes the red part was you moving the denominator to the numerator. Thanks so much and sorry about the missing parentheses. If I knew you, I would buy you a drink :p
• Jun 7th 2008, 12:47 PM
Moo
Quote:

Originally Posted by PensFan10
Yes the red part was you moving the denominator to the numerator. Thanks so much and sorry about the missing parentheses. If I knew you, I would buy you a drink :p

Great, I was being thirsty ! (Rofl)
• Jun 7th 2008, 12:48 PM
Mathstud28
Quote:

Originally Posted by PensFan10
First, I really need help with implicit differentiation. I tried reading in my book, but they use very basic problems and I did not find it helpful.
The problem I need help with is:

x^2 + xy - y^2 = 4 Solver for y'
I can not get the correct answer. I got (2x)/(x-2y)=y' (This is incorrect).

Next I need help with the chain rule. I understand how to do this, but I am having problems with one question. The question is:

Sqrt (2x)/(x^2+1.8)) <- For this problem, everything is under the radical.

Thank you!!!

Alternatively for the second one, Let

$y=\sqrt{\frac{2x}{x^2+1.8}}\Rightarrow{y^2=\frac{2 x}{x^2+1.8}}$

So differentiating we get

$2y\cdot{y'}=\frac{2(x^2+1.8)-(2x)(2x)}{(x^2+1.8)^2}=\frac{-2(x^2-1.8)}{(x^2+1.8)^2}\Rightarrow{y'=\frac{-(x^2-1.8)}{y(x^2+1.8)^2}}$

and now remembering that y is our original equation we see

$y'=\frac{-(x^2-1.8)}{\sqrt{\frac{2x}{x^2+1.8}}(x^2+1.8)^2}$

EDIT:

Same concept, just a little less messy to write
• Jun 7th 2008, 01:09 PM
PensFan10
okay
i kind of see what you did. I understand squaring both sides and then taking the derivative. I got lost after you simplified.

how did you go from 2(x^2+1.8)-(2x)(2x) to -2(x^2-1.8) (i left out the denominator to be less confusing)

Second, is that actually supposed to be a 4.8 in the numerator in your final answer or is it a typo. *EDIT* never mind. i see you changed it. now that part makes more sense :P
• Jun 7th 2008, 01:11 PM
Mathstud28
Quote:

Originally Posted by PensFan10
i kind of see what you did. I understand squaring both sides and then taking the derivative. I got lost after you simplified.

how did you go from 2(x^2+1.8)-(2x)(2x) to -2(x^2-1.8) (i left out the denominator to be less confusing)

Second, is that actually supposed to be a 4.8 in the numerator in your final answer or is it a typo.

That is a typo that I fixed, and secondly

$(2(x^2+1.8)-(2x)(2x)=2x^2+2(1.8)-4x^2=-2x^2-(-2)(1.8)=-2(x^2-1.8)$
• Jun 7th 2008, 01:16 PM
PensFan10
haha
Ha. Not sure how I didn't see that (Headbang) Thanks. I really like that way.