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

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To differentiate $\cos(x+y)$ given the equation $y^2+\cos(x+y) = 1$, the chain rule is applied. The derivative is expressed as $-\sin(x+y)(1+\frac{dy}{dx})$. This leads to the equation $2y\frac{dy}{dx}-\sin(x+y)(1+\frac{dy}{dx})=0$. Solving this equation allows for the determination of $\frac{dy}{dx}$. The discussion emphasizes the correct application of differentiation techniques.
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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
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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