How do I differentiate $\cos(x+y)$?

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SUMMARY

The discussion focuses on differentiating the expression $\cos(x+y)$ using implicit differentiation. Given the equation $y^2 + \cos(x+y) = 1$, the chain rule is applied, resulting in the derivative $\frac{dy}{dx}$. The final equation derived is $2y\frac{dy}{dx} - \sin(x+y)(1 + \frac{dy}{dx}) = 0$, which can be solved for $\frac{dy}{dx}$ to find the relationship between $x$ and $y$.

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If $y^2+\cos(x+y) = 1$ find $\frac{dy}{dx}$. How do I differentiate $\cos(x+y)$ bit?
 
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Guest said:
If $y^2+\cos(x+y) = 1$ find $\frac{dy}{dx}$. How do I differentiate $\cos(x+y)$ bit?

Use the chain rule:

$$\frac{d}{dx}\left(\cos(u(x))\right)=-\sin(u(x))\d{u}{x}$$

If $u(x)=x+y$, then what do you get? :)
 
MarkFL said:
Use the chain rule:

$$\frac{d}{dx}\left(\cos(u(x))\right)=-\sin(u(x))\d{u}{x}$$

If $u(x)=x+y$, then what do you get? :)
I get $-\sin(x+y)(1+\frac{dy}{dx})$
 
Guest said:
I get $-\sin(x+y)(1+\frac{dy}{dx})$

Yes, that's correct. (Yes)

So, then you have:

$$2y\d{y}{x}-\sin(x+y)\left(1+\d{y}{x}\right)=0$$

And you just need to solve for $$\d{y}{x}$$. :)
 
MarkFL said:
Yes, that's correct. (Yes)

So, then you have:

$$2y\d{y}{x}-\sin(x+y)\left(1+\d{y}{x}\right)=0$$

And you just need to solve for $$\d{y}{x}$$. :)
Done! Many thanks! :D
 

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