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

[Guest2]

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

 

[MarkFL]

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? :)

 

[Guest2]

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 $-\sin(x+y)(1+\frac{dy}{dx})$

 

[MarkFL]

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 $$\d{y}{x}$$. :)

 

[Guest2]

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