# Thread: Implicit Differation and Chain Rule

1. ## 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!!!

2. Hello,

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 :

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

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 =)

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

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

So calculate $\displaystyle u'(x)$ 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).

3. ## 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!!!

4. ## 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!

5. 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 $\displaystyle u'(x)$ ?
Because I doubt it is $\displaystyle \frac 2x$ /:

Can you show your working so that I can tell you where the mistake(s) is(are) ?

6. ## 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.

7. 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 $\displaystyle -2x^2+3.6=2(-x^2+1.8)$ and simplify the fraction by 2

The solution is :

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

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

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

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

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

Now it depends on how you want to display the result

8. ## 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

9. 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
Great, I was being thirsty !

10. 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

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

So differentiating we get

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

and now remembering that y is our original equation we see

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

EDIT:

Same concept, just a little less messy to write

11. ## 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

12. 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

$\displaystyle (2(x^2+1.8)-(2x)(2x)=2x^2+2(1.8)-4x^2=-2x^2-(-2)(1.8)=-2(x^2-1.8)$

13. ## haha

Ha. Not sure how I didn't see that Thanks. I really like that way.