How do you solve for x when using the chain rule and functions with variables?

  • Thread starter cogs24
  • Start date
In summary, the person is asking how to integrate y=4-x^5x using the differentiation of a first order ODE. They were given the correct steps, but were asked to differentiate the function on the LHS and forgot to include the product rule. They were eventually able to solve the problem.
  • #1
cogs24
30
0
y = (4-x) ^ 5x

Just wondering what i should do with this
Thanx
 
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  • #2
I guess that depends what the question is. Are you asking for the differential equation to which this is the solution?

Take logs and differentiate for first order ODE?
 
  • #3
yes, i have to differentiate it

sorry i didnt specifiy what had to be done, i presumed you knew it was differentiation.
 
  • #4
If it's just finding the derivative then this is probably more a question for the calculus thread.

Anyhoo, i'll do the first part. Take logs to get.

[tex] ln(y)= 5x ln(4-x) [/tex]

Now differentiate (implicitly on the LHS) and rearrange to get your answer for dy/dx. Have a go and let me know if/where you get stuck.
 
  • #5
ok, i understand your instructions, its just a matter of doing the right things now
This is what i got as an answer, unfortunately we arent supplied with answers in this exercise.

following on from your step, this is what i did

1/y * dy/dx = 5 * (1/4-x)
dy/dx = (5/4-x)y
dy/dx = (5/4-x)(4-x)^5x
 
  • #6
Pay attention with the differentiaition of the RHS.It's a product.I'm sure one of the 2 terms will contain the natural logarithm.

Daniel.
 
  • #7
Think with my poor Tex you missed the x after the 5 on the RHS. Take that into account and you're there.
 
  • #8
cogs24 said:
y = (4-x) ^ 5x

Just wondering what i should do with this
Thanx
Here's another way:
Let [tex]y(x)=f(u(x),v(x)), f(u,v)=u^{v}, u(x)=4-x, v(x)=5x[/tex]
Then, we have:
[tex]\frac{dy}{dx}=\frac{\partial{f}}{\partial{u}}\frac{du}{dx}+\frac{\partial{f}}{\partial{v}}\frac{dv}{dx}[/tex]
 
  • #9
ahh i see. i didnt spot the product rule on the right, i guess practice makes perfect.
Thanx everyone for the input.
 
  • #10
when differentiating functions of a variable say x raised to another function of xthen let's assume they are T^u,where t and u are functions of x,dy/dx =
T^u[du/dx*logt+u(dt/dx)/t]
 
  • #11
when differentiating functions of a variable say x raised to another function of xthen let's assume they are T^u,where t and u are functions of x,dy/dx =
T^u[du/dx*logt+u(dt/dx)/t].now let 4-x=t,and u=5x.
take normal procedures and see if it works
 

Related to How do you solve for x when using the chain rule and functions with variables?

1. What is the value of x in the equation y = (4-x)^5?

The value of x cannot be determined without more information. This equation has two variables (x and y) and only one equation, so there are infinitely many solutions.

2. How can I solve for x in this equation?

To solve for x, you would need to have another equation or some additional information about the relationship between x and y. With just one equation, it is not possible to solve for a specific value of x.

3. Can I graph this equation?

Yes, you can graph this equation by choosing different values for x and finding the corresponding y values. However, since there are infinitely many solutions, the graph will not be a single line, but rather a curve.

4. Is there a specific method for solving this type of equation?

This equation does not fall into a specific category of equations that have a set method for solving. Each equation is unique and may require different techniques or additional information to find a solution.

5. Can I use a calculator to solve this equation?

No, a calculator cannot solve this equation for a specific value of x. It can only give you a numerical approximation for a given value of y. To find a specific value of x, you would need to use algebraic methods or have additional information about the relationship between x and y.

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