# Thread: Implict differentiation: 2cosysinx=1

1. ## Implict differentiation: 2cosysinx=1

Hello,

The question is to find the equation of the normal to the curve $\displaystyle 2\cos y\sin x = 1$ at $\displaystyle (\tfrac{\pi}{4},\tfrac{\pi}{4})$.

I'm not confident with implicit differentiation, but using the product rule I got:

$\displaystyle 2\sin x\dfrac{dy}{dx}(-\sin y) + 2\cos y\cos x = 0$

$\displaystyle \dfrac{dy}{dx}(2\sin x\sin y) - 2\cos y\cos x = 0$

$\displaystyle \dfrac{dy}{dx} = \dfrac{2\cos y\cos x}{2\sin x\sin y}$

$\displaystyle \dfrac{dy}{dx} = \cot x\cot y$

My question is, is it correct to write the derivative of $\displaystyle \cos y$ with respect to $\displaystyle x$ as $\displaystyle \dfrac{dy}{dx}(-\sin y)$ as I have done.

Thank you.

2. ## Re: Implict differentiation: 2cosysinx=1

Hey Furyan.

Your working looks good for the implicit differentiation.

3. ## Re: Implict differentiation: 2cosysinx=1

Hey chiro,

That's a relief, thank you very much for looking it over

### 2cosysinx

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