Trig Derivative Problem

[Jason76]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\sin(x) }{2x}\)

What is the first step?

The answer is \(\displaystyle \dfrac{x \cos(x) - \sin(x)}{2x^{2}} dx\)

 

[tkhunny]

Why is this different from any other Quotient Rule? Let's see you step through it.

Note: The answer appears to be correct.

 

[MarkFL]

The (possible) first step is to apply the quotient rule:

\(\displaystyle \displaystyle \frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}\)

 

[Deleted member 4993]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\sin(x) }{2x}\)

What is the first step?

The answer is \(\displaystyle \dfrac{x \cos(x) - \sin(x)}{2x^{2}} dx\)

The post above does not make sense - unless it is an antiderivative problem or ODE problem.

What are you supposed to find? first step to find what? Is there a definition of 'y' somewhere?

 

[lookagain]

\(\displaystyle \dfrac{dy}{dx} = \dfrac{\sin(x) }{2x}\)

What is the first step?

The answer is \(\displaystyle \dfrac{x \cos(x) - \sin(x)}{2x^{2}} dx\)

No, Jason76, the problem should be:

\(\displaystyle y \ = \ \dfrac{\sin(x)}{2x}\)

Determine \(\displaystyle \ \dfrac{dy}{dx}.\)



So start by applying the hints/comments from tkhunny and MarkFL.


If the problem wants \(\displaystyle \dfrac{dy}{dx}, \ \ \)then there will be no "dx" in the answer.

If the problem wants dy, then there would be a "dx" as part of the answer.





Please type the problem in full.