# Josh's question at Yahoo! Answers regarding implicit differentiation

#### MarkFL

Staff member
Here is the question:

Derive Trig Function?

2. 2x + y2 – sin(xy) = 0

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

Staff member
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.