Implicit Differentiation Problem - Check my work?

In summary, the conversation involves someone asking for their work to be checked on a problem involving implicit differentiation and another person providing a summary of the steps taken to solve the problem and offering help with a different problem.
  • #1
CACain
6
0
Implicit Differentiation Problem -- Check my work?

I've worked it -- can someone just check my work?

Problem:

xcosy+ycos=1

My work:

[x (d/x)cosy + cosy (d/dx)x] + [y (d/dx)cosx + cosx (d/dx)y] = (d/dx) 1

-xsiny (dy/dx) + cos y - ysinx + cos x (dy/dx) = 0

-xsiny (dy/dx) + cos y = ysinx - cosy

dy/dx = (ysinx - cosy)/(-xsiny + cos x)


Meanwhile, could someone help me with this one...

squareroot (xy) = 1+(x^2)y
 
Last edited:
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  • #2
the first one looks good, as for the second:



(2)[tex]xy=1+x^2y[/tex]

You are differentiating WRT y, which means that the derivative of x is 1, but the derivative of y is dy/dx. Make sure you use the product rule:

[tex]x\frac{dy}{dx}+y=x^2 \frac{dy}{dx}+2xy[/tex]

[tex]x\frac{dy}{dx}-x^2 \frac{dy}{dx}=2xy-y[/tex]

[tex]\frac{dy}{dx}(x-x^2)=2xy-y[/tex]

[tex]\frac{dy}{dx}=\frac{2xy-y}{x-x^2}[/tex]

~Josh
 
  • #3
-xsiny (dy/dx) + cos y = ysinx - cosy

dy/dx = (ysinx - cosy)/(-xsiny + cos x)

You miswrote the first line above but obviously that was a typo since you got it right in the end. If I were your teacher I would prefer to see one more line:
-xsiny (dy/dx) + cos y = ysinx - cos x(dy/dx)

cos x (dy/dx)- x sin y (dy/dx)= (cos x- x sin y)(dy/dx)= y sin x- cos y

dy/dx = (ysinx - cosy)/(-xsiny + cos x)
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of an implicit equation, where the dependent variable is not explicitly written in terms of the independent variable.

2. How do I know when to use implicit differentiation?

Implicit differentiation is typically used when an equation is given in the form of f(x,y) = 0, and the dependent variable y is not explicitly solved for. It is also used when taking the derivative of inverse trigonometric functions.

3. What are the steps involved in implicit differentiation?

The steps involved in implicit differentiation are as follows:1. Differentiate both sides of the equation with respect to the independent variable.2. For any terms with the dependent variable y, use the chain rule and multiply by dy/dx.3. Group all terms with dy/dx on one side of the equation.4. Solve for dy/dx.

4. How can I check my work when using implicit differentiation?

To check your work when using implicit differentiation, you can substitute your calculated value for dy/dx back into the original equation. If the resulting equation is true, then you have successfully found the derivative using implicit differentiation.

5. Are there any common mistakes to avoid when using implicit differentiation?

Yes, there are a few common mistakes to avoid when using implicit differentiation. These include:- Forgetting to use the chain rule when differentiating terms with y.- Incorrectly solving for dy/dx and leaving out a factor of dx.- Not simplifying the resulting equation.- Mistakes in basic algebraic manipulations.It is important to carefully check your work and be aware of these potential errors.

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