differentiation?

[sdracula]

hint does not work for this. i have knowledge but no progress for the answer.

 

[tkhunny]

What do you get for the derivative after applying the technique of "Implicit Differentiation"?

Hint: Using xy = 3 ==> Implicit Differentiation using the Product Rule ==> xy' + y = 0 ==> Using Algebra ==> y' = -y/x

 

[sdracula]

What do you get for the derivative after applying the technique of "Implicit Differentiation"?

Hint: Using xy = 3 ==> Implicit Differentiation using the Product Rule ==> xy' + y = 0 ==> Using Algebra ==> y' = -y/x

as I said, i know rules but what does the question want

 

[MarkFL]

Hello, and welcome to FMH!

We are given:

[MATH]y+x\tan(ky)=0[/MATH]

Spoiler

Implicitly differentiate w.r.t \(x\)

[MATH]\d{y}{x}+\tan(ky)+x\sec^2(ky)\left(k\d{y}{x}\right)=0[/MATH]
Solve for [MATH]\d{y}{x}[/MATH]:

[MATH]\d{y}{x}=-\frac{\tan(ky)}{1+kx\sec^2(ky)}[/MATH]
Apply Pythagorean identity [MATH]\tan^2(\theta)+1=\sec^2(\theta)[/MATH]:

[MATH]\d{y}{x}=-\frac{\tan(ky)}{1+kx(\tan^2(ky)+1)}[/MATH]
From the given equation, we know:

[MATH]\tan(ky)=-\frac{y}{x}[/MATH]
Thus:

[MATH]\d{y}{x}=-\frac{-\dfrac{y}{x}}{1+kx\left(\left(-\dfrac{y}{x}\right)^2+1\right)}=\frac{y}{k(x^2+y^2)+x}[/MATH]
Using the small angle approximation, the original equation becomes:

[MATH]y+kxy=0[/MATH]
[MATH]1+kx=0[/MATH]
[MATH]x=-\frac{1}{k}[/MATH]

 

[tkhunny]

as I said, i know rules but what does the question want

That's no good. You were asked a question. Answer it. What do YOU get? This was the question. This is where you should have started the conversation. Indeed, Welcome to FMH. There is WORK to do and to show.

 

[sdracula]

That's no good. You were asked a question. Answer it. What do YOU get? This was the question. This is where you should have started the conversation. Indeed, Welcome to FMH. There is WORK to do and to show.

sorry, i am new in forum. I can't focus enough because of my limited time. I feel like a zombie at now. look at the my time zone now (GMT+3). If I didn't need it, i would definitely deal with it myself. I will ask you one more question tomorrow. This time we will do it together as you said. I hope you can help. by the way implicit dif. means to me ''importance of dependent variable etc.'' do you think we could go forward from here. at the beginning-the function seem complicated.

 

[pka]

If \(\displaystyle y+x\tan(ky)=0\) then:
\(\displaystyle y'+\tan(ky)+x\sec^2(ky)\cdot(ky')=0\)

Now you solve for \(\displaystyle y'\) and use trigonometric identities to simplify.

 

[sdracula]

thank you FMH, solved.