Josh's question at Yahoo Answers regarding implicit differentiation

[MarkFL]

Here is the question:

Derive Trig Function?

2. 2x + y2 – sin(xy) = 0

Please help me derive this, thanks!Additional Details

the equation is 2x + y^2 - sin(xy) = 0 The y is SQUARED

Here is a link to the question:

Derive Trig Function? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Josh,

We are given to find \(\displaystyle \frac{dy}{dx}\) for the implicit relation:

\(\displaystyle 2x+y^2-\sin(xy)=0\)

So, we may use implicit differentiation since the equation cannot be solved for either $x$ or $y$. This involves using the chain rule, along with the other rules for differentiation.

Hence, we find by implicitly differentiating with respect to $x$:

\(\displaystyle 2+2y\frac{dy}{dx}-\cos(xy)\frac{d}{dx}(xy)=0\)

Applying the product rule, there results:

\(\displaystyle 2+2y\frac{dy}{dx}-\cos(xy)\left(x\frac{dy}{dx}+y \right)=0\)

Distributing the cosine function, we have:

\(\displaystyle 2+2y\frac{dy}{dx}-x\frac{dy}{dx}\cos(xy)-y\cos(xy)=0\)

Now, we move all terms not having \(\displaystyle \frac{dy}{dx}\) as a factor to the right side, and factor out \(\displaystyle \frac{dy}{dx}\) on the left side:

\(\displaystyle \frac{dy}{dx}\left(2y-x\cos(xy) \right)=y\cos(xy)-2\)

Divide through by \(\displaystyle 2y-x\cos(xy)\) to obtain:

\(\displaystyle \frac{dy}{dx}=\frac{y\cos(xy)-2}{2y-x\cos(xy)}\)

To Josh and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.