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

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In summary: ing over an equation will give you information about it, such as its type and the variables involved-in equation-type mode (default key is "c"), you can type "t" to activate the tangent function and "s" to activate the secant function
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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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  • #2
Your last line is correct. Perhaps the problem is with how you entered it. What is the exact string that you entered?
 
  • #3
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?
 
  • #4
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.
 
  • #5
Ok. Thanks for double checking for me then.
 
  • #6
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
 

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly expressed in terms of one variable. This allows us to find the rate of change of a function at a specific point, even when the function is not in the form y = f(x).

Why do we use implicit differentiation?

Implicit differentiation is useful when we have a function that is difficult or impossible to express explicitly in terms of one variable. It allows us to find the derivative of the function and use it to solve problems involving rates of change and optimization.

How do you perform implicit differentiation?

To perform implicit differentiation, we use the chain rule and the product rule to differentiate each term in the equation with respect to the variable of interest. We then solve for the derivative of the function by isolating the variable on one side of the equation.

What is the difference between implicit differentiation and explicit differentiation?

Explicit differentiation involves finding the derivative of a function that is expressed explicitly in terms of one variable. Implicit differentiation, on the other hand, is used when the function is not expressed explicitly in terms of one variable, but rather as an equation involving multiple variables.

What are some real-life applications of implicit differentiation?

Implicit differentiation has many real-life applications, such as in physics, economics, and engineering. For example, it can be used to find the rate of change of a population over time, the maximum profit for a company, or the optimal shape for a bridge design.

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