# Implict differentiation: 2cosysinx=1

• Jan 26th 2013, 06:27 PM
Furyan
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.
• Jan 26th 2013, 06:31 PM
chiro
Re: Implict differentiation: 2cosysinx=1
Hey Furyan.

Your working looks good for the implicit differentiation.
• Jan 26th 2013, 06:39 PM
Furyan
Re: Implict differentiation: 2cosysinx=1
Hey chiro,

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