Implicit differentiation homework

In summary, A question is asked about implicit differentiation, and after some discussion and confusion, it is suggested to use substitution to solve the problem. The final answer is provided and one person comments that it is a better method.
  • #1
UNknown 2010
77
0
hi

i couldn't solve this question
i think there is a mistake in it
anyone can check it please ?
http://img296.imageshack.us/img296/3716/13055173zr3.png
 
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  • #2


Use implicit differentiation:
The derivative of 5y, with respect to x, is 5 dy/dx and the derivative of 5xy, with respect to x, is 5y+ 5x dy/dx.

Differentiate both sides of the equation with respect to x and solve for dy/dx.
 
  • #3


http://img394.imageshack.us/img394/7196/94146936ef8.png
then I got stuck
 
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  • #4


Now solve your original equation for y and substitute it into your result.
 
  • #5
UNknown 2010 said:
dy/dx = (5y - 2)/-5(x+1)
then I got stuck

And 5y = … ? :smile:
 
  • #6


5y=2x/(x+1)
 
  • #7
JANm said:
5y=2x/(x+1)

Yup! :smile:

So dy/dx = (5y - 2)/-5(x+1) = … ? :wink:
 
  • #8


Tiny Tim
Ever heard of substitution: instead of 5y in the differential equation you may substitute
2x/(x+1) while the first equation seems to be right.
 
  • #9
JANm said:
Tiny Tim
Ever heard of substitution: instead of 5y in the differential equation you may substitute
2x/(x+1) while the first equation seems to be right.

Sorry, not following you. :confused:

btw, are you the same person as UNknown 2010?
 
  • #10


No I am knot!
 
  • #11


JANm said:
No I am knot!

You're "a method for fastening or securing linear material such as rope by tying or interweaving."? :confused:\

Disclaimer: (1) Comment said in jest. (2) Quote is the definition of the term 'knot', taken from wiki :smile:
 
  • #12


I don't know if you might prefer this method, but this is how I would've done it since I don't have a clue about what everyone else has suggested:

[tex]2x-5y=5xy[/tex]

[tex]y(5x+5)=2x[/tex]

[tex]y=\frac{2x}{5(x+1)}[/tex]

Now just use the quotient rule. [tex]\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex] where u=numerator, v=denominator.

EDIT: Felt like I was giving away the answer (literally)
 
Last edited:
  • #13
Mentallic said:
I don't know if you might prefer this method, but this is how I would've done it since I don't have a clue about what everyone else has suggested …

Hi Mentallic! :smile:

I think your way is better! :biggrin:
 

FAQ: Implicit differentiation homework

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly expressed in terms of the independent variable. It involves treating the dependent variable as a function of the independent variable and using the chain rule to find the derivative.

2. When is implicit differentiation used?

Implicit differentiation is typically used when the function cannot be easily solved for the dependent variable. It is also useful when the function is expressed as an equation rather than a function.

3. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves solving the function for the dependent variable and then taking the derivative. Implicit differentiation, on the other hand, involves treating the dependent variable as a function of the independent variable and using the chain rule to find the derivative.

4. What are the steps for performing implicit differentiation?

The steps for implicit differentiation are as follows:

  1. Identify the dependent and independent variables in the function.
  2. Differentiate both sides of the equation with respect to the independent variable.
  3. Use the chain rule to find the derivative of the dependent variable.
  4. Solve for the derivative of the dependent variable.

5. Can implicit differentiation be used for any type of function?

Yes, implicit differentiation can be used for any type of function as long as the dependent variable is expressed as a function of the independent variable in the equation.

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