Implicit differentiation(Help ASAP!)

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Zerrotolerance

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I am having an extremely difficult time understanding this subject and what to do. I think I have the chain, product, and quotient rule for derivitives down pretty well, but when I combine them to solve difficult problems I am losing track of what to do when and how. Can someone please explain how to do this problem for me step by step so I get a better idea. It doesn't need to be simplified just find the derivitive.

y[sup:3z6s71kp]5[/sup:3z6s71kp]+ x[sup:3z6s71kp]5[/sup:3z6s71kp] y[sup:3z6s71kp]3[/sup:3z6s71kp]= -9 + y e[sup:3z6s71kp]x[/sup:3z6s71kp][sup:3z6s71kp]2[/sup:3z6s71kp]

*****for some reason the problem isn't showing up correctly. The last part should be y times e raised to the x squared or ye^x^2.*******

This is what I have so far:

1) Find d/dx of both sides

This gives me:

chain rule chain rule and product rule
5y[sup:3z6s71kp]4[/sup:3z6s71kp](dy/dx) + (x[sup:3z6s71kp]5[/sup:3z6s71kp]3y[sup:3z6s71kp]2[/sup:3z6s71kp](dy/dx) + 5x[sup:3z6s71kp]4[/sup:3z6s71kp]y[sup:3z6s71kp]3[/sup:3z6s71kp]) = 2xe[sup:3z6s71kp]x[/sup:3z6s71kp][sup:3z6s71kp]2[/sup:3z6s71kp]y

**I can't figure out how to get the derivitive of the ye[sup:3z6s71kp]x[/sup:3z6s71kp][sup:3z6s71kp]2[/sup:3z6s71kp]. I know you use the chain and product rule, but that's it. I used my calculator to get the answer and I don't know if it's right.

2) get (dy/dx) alone

dy/dx = ((2xe[sup:3z6s71kp]x[/sup:3z6s71kp][sup:3z6s71kp]2[/sup:3z6s71kp]-5x[sup:3z6s71kp]4[/sup:3z6s71kp]y[sup:3z6s71kp]3[/sup:3z6s71kp])/(5y[sup:3z6s71kp]4[/sup:3z6s71kp] + x[sup:3z6s71kp]5[/sup:3z6s71kp]3y[sup:3z6s71kp]2[/sup:3z6s71kp])

I'm getting marked wrong on this but I don't know where. I skipped a few steps, but I don't think it's my algebra that's messed up by getting dy/dx alone.

Can someone point out how to correctly solve this and how to get the derivitive above I couldn't find? Thanks!
 
Zerrotolerance said:
I am having an extremely difficult time understanding this subject and what to do. I think I have the chain, product, and quotient rule for derivitives down pretty well, but when I combine them to solve difficult problems I am losing track of what to do when and how. Can someone please explain how to do this problem for me step by step so I get a better idea. It doesn't need to be simplified just find the derivitive.

y[sup:1g686po4]5[/sup:1g686po4]+ x[sup:1g686po4]5[/sup:1g686po4] y[sup:1g686po4]3[/sup:1g686po4]= -9 + y e[sup:1g686po4]x[/sup:1g686po4][sup:1g686po4]2[/sup:1g686po4]

*****for some reason the problem isn't showing up correctly. The last part should be y times e raised to the x squared or ye^x^2.*******

This is what I have so far:

1) Find d/dx of both sides

This gives me:

chain rule chain rule and product rule
5y[sup:1g686po4]4[/sup:1g686po4](dy/dx) + (x[sup:1g686po4]5[/sup:1g686po4]3y[sup:1g686po4]2[/sup:1g686po4](dy/dx) + 5x[sup:1g686po4]4[/sup:1g686po4]y[sup:1g686po4]3[/sup:1g686po4]) = 2xe[sup:1g686po4]x[/sup:1g686po4][sup:1g686po4]2[/sup:1g686po4]y + e[sup:1g686po4]x[/sup:1g686po4][sup:1g686po4]2[/sup:1g686po4]*(dy/dx)

**I can't figure out how to get the derivitive of the ye[sup:1g686po4]x[/sup:1g686po4][sup:1g686po4]2[/sup:1g686po4]. I know you use the chain and product rule, but that's it. I used my calculator to get the answer and I don't know if it's right.

2) get (dy/dx) alone

dy/dx = ((2xe[sup:1g686po4]x[/sup:1g686po4][sup:1g686po4]2[/sup:1g686po4]-5x[sup:1g686po4]4[/sup:1g686po4]y[sup:1g686po4]3[/sup:1g686po4])/(5y[sup:1g686po4]4[/sup:1g686po4] + x[sup:1g686po4]5[/sup:1g686po4]3y[sup:1g686po4]2[/sup:1g686po4])

I'm getting marked wrong on this but I don't know where. I skipped a few steps, but I don't think it's my algebra that's messed up by getting dy/dx alone.

Can someone point out how to correctly solve this and how to get the derivitive above I couldn't find? Thanks!
Click to expand...
 
I just discovered that it was the derivitive of ye^x^2 that screwed the entire thing up. Besides that I have everything else correctly setup. I think I understand now. I wonder why my calculator got it wrong?

ye^x^2

so,

if you were to rewrite this as:

f(x)=y
f'(x) = 1(dy/dx)
g(x)=e^x^2
g'(x)=2xe^x^2

then.

(y)(2xe^x^2)+(dy/dx)(e^x^2) = 2xe^x^2*y+e^x^2(dy/dx)

I forgot to do the product rule.....

I am having a very hard time knowing when to use which rules even though I know how to use the rules. Can anyone give me some advice. I start seeing complitcated functions like y = 5^3^x^2 and I freak out. Now that I'm at implicit functions I am having an even more difficult time. I'll probably be back soon with another question, but for now I am ok. Thank you for the fast reply and help Khan, I would be lost without this site.
 
\(\displaystyle Let \ f \ = \ [y^5+x^5y^3 \ = \ -9+ye^{x^2}].\)

\(\displaystyle Then \ implicit \ diff. \ (f,y,x) \ = \ 5y^4y'+5x^4y^3+3x^5y^2y' \ = \ 0+y'e^{x^2}+2xye^{x^2}.\)

\(\displaystyle 5y^4y'+3x^5y^2y'-y'e^{x^2} \ = \ 2xye^{x^2}-5x^4y^3.\)

\(\displaystyle Hence, \ y' \ = \ \frac{dy}{dx} \ = \ \frac{xy(2e^{x^2}-5x^3y^2)}{5y^4+3x^5y^2-e^{x^2}}\)

\(\displaystyle implicit \ diff. \ (f,x,y) \ = \ 5y^4+5x^4y^3x'+3x^5y^2 \ = \ 0+e^{x^2}+2xyx'e^{x^2}.\)

\(\displaystyle 5x^4y^3x'-2xyx'e^{x^2} \ = \ e^{x^2}-5y^4-3x^5y^2.\)

\(\displaystyle Therefore, \ x' \ = \ \frac{dx}{dy} \ = \ \frac{e^{x^2}-5y^4-3x^5y^2}{xy(5x^3y^2-2e^{x^2})}.\)
 
Thank you. I finally have a good understanding of this material. I just needed to see it out and try it a few times. I just finished 10 more similar problems relatively easy. I am struggling with one last question on this HW assignment. I did the hard part, but am having trouble simplifying. (Part a) is complete and correct, it's part b) giving me difficulties because I am not sure what it wants.

The problem is:

(a) Find the second derivative y '' by implicit differentiation if:

x[sup:5bzsv96i]3[/sup:5bzsv96i]+ y[sup:5bzsv96i]3[/sup:5bzsv96i]= 18

Express your answer in terms of x and y alone (so that y ' does not appear in your answer).
Do not solve for y in terms of x.

a) y '' = (-2xy[sup:5bzsv96i]2[/sup:5bzsv96i]+2x[sup:5bzsv96i]2[/sup:5bzsv96i]y(-x[sup:5bzsv96i]2[/sup:5bzsv96i]/y[sup:5bzsv96i]2[/sup:5bzsv96i]))/y[sup:5bzsv96i]4[/sup:5bzsv96i]

(b) Now simplify your answer to (a) as much as possible, still without solving for y in terms of x. (Hint: For the final simplification, use the relation between x and y given by the original equation.)

b) y '' =

P.S. How do you guys show your work out, so you get division and square root to actually appear. I think it would be much easier to post and for you to understand what i mean if i knew how. Do you just do it in another program and place it here as an image? Is there certain code to do it because I don't see it?

Thanks and sorry for all the questions!
 
Zerrotolerance said:
I wonder why my calculator got it wrong?

If you would like to know, show us what you entered and the calculator's response.
Click to expand...
 
Zerrotolerance said:
Thank you. I finally have a good understanding of this material. I just needed to see it out and try it a few times. I just finished 10 more similar problems relatively easy. I am struggling with one last question on this HW assignment. I did the hard part, but am having trouble simplifying. (Part a) is complete and correct, it's part b) giving me difficulties because I am not sure what it wants.

The problem is:

(a) Find the second derivative y '' by implicit differentiation if:

x[sup:2f4opk6f]3[/sup:2f4opk6f]+ y[sup:2f4opk6f]3[/sup:2f4opk6f]= 18

Express your answer in terms of x and y alone (so that y ' does not appear in your answer).
Do not solve for y in terms of x.

a) y '' = (-2xy[sup:2f4opk6f]2[/sup:2f4opk6f]+2x[sup:2f4opk6f]2[/sup:2f4opk6f]y(-x[sup:2f4opk6f]2[/sup:2f4opk6f]/y[sup:2f4opk6f]2[/sup:2f4opk6f]))/y[sup:2f4opk6f]4[/sup:2f4opk6f]

(b) Now simplify your answer to (a) as much as possible, still without solving for y in terms of x. (Hint: For the final simplification, use the relation between x and y given by the original equation.)

b) y '' =

P.S. How do you guys show your work out, so you get division and square root to actually appear. I think it would be much easier to post and for you to understand what i mean if i knew how. Do you just do it in another program and place it here as an image? Is there certain code to do it because I don't see it?

Thanks and sorry for all the questions!
Click to expand...

This is simple algebra - just needs careful manipulation of symbols - no tricks involved

a) y '' = (-2xy[sup:2f4opk6f]2[/sup:2f4opk6f]+2x[sup:2f4opk6f]2[/sup:2f4opk6f]y(-x[sup:2f4opk6f]2[/sup:2f4opk6f]/y[sup:2f4opk6f]2[/sup:2f4opk6f]))/y[sup:2f4opk6f]4[/sup:2f4opk6f]

= (-2xy[sup:2f4opk6f]2[/sup:2f4opk6f]-2x[sup:2f4opk6f]4[/sup:2f4opk6f]/y)/y[sup:2f4opk6f]4[/sup:2f4opk6f]

= -2(xy[sup:2f4opk6f]3[/sup:2f4opk6f] + x[sup:2f4opk6f]4[/sup:2f4opk6f])/y[sup:2f4opk6f]5[/sup:2f4opk6f]

Can you finish it now....
 
Zerrotolerance said:
y '' = (-2xy[sup:3ggsizpg]2[/sup:3ggsizpg]+2x[sup:3ggsizpg]2[/sup:3ggsizpg]y(-x[sup:3ggsizpg]2[/sup:3ggsizpg]/y[sup:3ggsizpg]2[/sup:3ggsizpg]))/y[sup:3ggsizpg]4[/sup:3ggsizpg]


This simplifies to -2x(x^3 + y^3)/y^5

You can now simplify it further, for part (b), yes ?


P.S. How do you guys show your work out, so you get division and square root to actually appear.

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