# Verification of Implicit Differentiation

1. Jan 18, 2005

Help with Implicit Differentiation

Hello all

If we are given $$\cos xy = 2x^2 - 3y$$ find $$\frac {dy}{dx}$$

So the derivative of $$\cos xy$$ is $$- sin(xy)(x)(\frac{dy}{dx} + y)$$

The derivative of the RHS is $$4x - 3 \frac {dy}{dx}$$

Hence $$\-sin(xy)(x)\frac{dy}{dx} + y$$ = $$4x - 3 \frac {dy}{dx}$$

HOw do I find $$\frac {dy}{dx}$$ ?

Is the answer $$\frac {dy}{dx} = \frac{y\sin xy + 4x}{ -x\sin xy + 3}$$

Thanks

( sin should be (- sin)

Last edited: Jan 18, 2005
2. Jan 18, 2005

### Yegor

I think You do not correctly found derivative of left part.
Hint: (x*y)'=y+x*y'

3. Jan 18, 2005

### Yegor

Sorry, I didn't noticed correction. You are right.

4. Jan 18, 2005

anybody who can verify if my answer is right?

5. Jan 18, 2005

### Yegor

The answer is right. You have tipos

6. Jan 18, 2005

### MathStudent

I think you have the right idea, but you need to clean up your work,,, for example
the fifth line should not be
$$-sin(xy)(x)\frac{dy}{dx} + y$$
you need to distribute the $\-sin(xy)$ to the y as well
so it should be:
$$-sin(xy)(x\frac{dy}{dx} + y)$$ ( I assumed you knew this, and just had difficulty with the syntax )

Anyway, once you have differentiated implicitly you can treat $\frac{dy}{dx}$ as a your unknown and solve for it algebraicly

Keep at it :)