MHB Differentiating a trignometric function

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The discussion revolves around differentiating a trigonometric function, specifically questioning the origin of the cosine term in the equation. The user is confused about how the transition from secant squared to cosine squared occurs in the differentiation process. A key point raised is the definition of secant, which is the reciprocal of cosine. This clarification helps to understand the relationship between secant and cosine in the context of differentiation. The conversation emphasizes the importance of recognizing trigonometric identities when differentiating functions.
tmt1
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So, I have

$$\d{x}{t} = 20 sex^2∂ \d{∂}{t}$$

And the text goes to:

$$ \d{∂}{t} = \frac{1}{20}cos^2 ∂ \d{x}{t}$$

I don't understand where the cos comes from? Is it a trigonometric identity? If so I can't find it.
 
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tmt said:
So, I have

$$\d{x}{t} = 20 sex^2∂ \d{∂}{t}$$

And the text goes to:

$$ \d{∂}{t} = \frac{1}{20}cos^2 ∂ \d{x}{t}$$

I don't understand where the cos comes from? Is it a trigonometric identity? If so I can't find it.

Hi tmt!

By definition:
$$\sec \delta = \frac{1}{\cos\delta}$$
 

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)}...
View full post »

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