Implicit Differentiation: How to Find dy/dx of an Equation with y and x Terms?

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Homework Help Overview

The discussion revolves around finding the derivative dy/dx of the equation y*sec(x) = 3x*tan(y). Participants explore the application of implicit differentiation and the use of various derivative rules.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the product rule and the derivatives of sec(x) and tan(y). There is a focus on the correctness of the final expression for dy/dx and the potential for input errors when submitting answers online.

Discussion Status

Some participants confirm the correctness of the original poster's last line of differentiation, while others suggest checking the input format for potential errors. The original poster later realizes a misunderstanding regarding the derivative requested, leading to a resolution of their confusion.

Contextual Notes

There is a mention of a misunderstanding regarding whether the derivative required was dy/dx or dx/dy, which impacted the original poster's approach to the problem.

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Homework Statement


Find dy/dx of this equation -
[tex]y*sec(x)=3x*tan(y)[/tex]


Homework Equations



-product rule
-derivative of sec(x) with respect to x is sec(x)tan(x) i believe
-derivative of tan(x) is sec^2(x) i believe


The Attempt at a Solution



[tex]y*sec(x)=3x*tan(y)[/tex]

[tex]y*sec(x)*tan(x)+sec(x) \frac{dy}{dx}=3x*sec^2(y)\frac{dy}{dx}+3*tan(y)[/tex]

[tex]sec(x)\frac{dy}{dx}-3x*sec^2(x)\frac{dy}{dx}=3*tan(y) -y*sec(x)*tan(x)[/tex]

[tex]\frac{dy}{dx}(sec(x)-3x*sec^2(x))= 3*tan(y) -y*sec(x)*tan(x)[/tex]

[tex]\frac{dy}{dx}= \frac{3*tan(y) -y*sec(x)*tan(x)}{sec(x)-3x*sec^2(y)}[/tex]

That last line is my solution. I do homework online and every time i enter this it says it is wrong. So where am I going wrong?
 
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Your last line is correct. Perhaps the problem is with how you entered it. What is the exact string that you entered?
 
slider142 said:
Your last line is correct. Perhaps the problem is with how you entered it. What is the exact string that you entered?

I entered it the almost the exact same way except for sec^2 . I typed it in as (sec(y))^2. Are you sure I am right?
 
hover said:
I entered it the almost the exact same way except for sec^2 . I typed it in as (sec(y))^2. Are you sure I am right?

That's the unambiguous way to write it. I have checked the differentiation independently. In your derivation, you switched one of the y's for an x, but you got the last line, so I assume that was a typo. If your parentheses are all correct in your typed version, I would go ahead and show the professor the problem and your solution personally.
 
Ok. Thanks for double checking for me then.
 
FIGURED IT OUT. The answer I have here is correct. Its a hundred percent correct. dy/dx does equal that value that i posted in my original post. TOO BAD THEY ASKED FOR dx/dy. Its the classic case of not reading the question correctly :P. I'm so used to taking the derivative of dy/dx I just made the assumption that it was dy/dx. Once I found out that it was dx/dy I was able to get the answer right on the first try :D. Sorry about causing you any trouble if I did.

Many thanks for your checking
-hover
 

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