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

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Discussion Overview

The discussion revolves around the differentiation of the expression $\cos(x+y)$ within the context of the equation $y^2+\cos(x+y) = 1$. Participants explore the application of the chain rule to find $\frac{dy}{dx}$, focusing on the differentiation process and the resulting expressions.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how to differentiate $\cos(x+y)$, indicating a need for clarification on the differentiation process.
  • Another participant suggests using the chain rule, stating that the derivative of $\cos(u(x))$ is $-\sin(u(x))\frac{du}{dx}$, where $u(x) = x + y$.
  • A subsequent reply confirms the application of the chain rule, resulting in the expression $-\sin(x+y)(1+\frac{dy}{dx})$.
  • Further, it is noted that this leads to the equation $2y\frac{dy}{dx}-\sin(x+y)(1+\frac{dy}{dx})=0$, which participants agree is correct.

Areas of Agreement / Disagreement

Participants generally agree on the application of the chain rule and the resulting expressions, but the discussion does not resolve the final steps for solving $\frac{dy}{dx}$.

Contextual Notes

The discussion does not address potential limitations or assumptions in the differentiation process, nor does it clarify the implications of the derived equations.

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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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