MHB Using Chain rule to find derivatives....

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To find the derivative of y = (csc(x) + cot(x))^-1, the chain rule is applied using u = csc(x) + cot(x). The derivative dy/dx is calculated as dy/dx = (dy/du)(du/dx). The discussion emphasizes the need to determine dy/du and du/dx, with a suggestion to simplify u using trigonometric identities. It is recommended to express u in terms of sine and cosine to facilitate the differentiation process, potentially employing the quotient rule for du/dx. Understanding these steps is crucial for successfully applying the chain rule in this context.
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y = (csc(x) + cot(x) )^-1
Find dy/dx
 
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Suppose u = csc(x) + cot x

dy/dx = dy/du × du/dx
 
darkknight said:
y = (csc(x) + cot(x) )^-1
Find dy/dx
darkknight, you titled this "using chain rule" and Monoxdifly just told you what that is! Are you able to do this now?

Letting u= csc(x)+ cot(x), y= u^-1. What is dy/du? What is du/dx?
chain rule: dy/dx= (dy/du)(du/dx).

If the difficulty is du/dx, it might help you to write [math]u= csc(x)+ cot(x)= \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}= \frac{1+ cos(x)}{sin(x)}[/math] and use the "quotient rule".
 

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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