MHB Derivatives of trigonometric functions

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The derivative of the function y=5tanx+4cotx is correctly identified as y'=5sec^2x-4csc^2x, which aligns with outputs from reliable math sources. The initial claim of y'=-5csc^2x is incorrect, as it does not hold as an identity when compared to the correct derivative. The discussion highlights the importance of verifying derivative calculations, especially before a calculus test. Clarification on the source of the incorrect result is sought, emphasizing the need for accurate mathematical understanding. Accurate differentiation is crucial for success in calculus.
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The answer for derivative of y=5tanx+4cotx is y'=-5cscx^2. But how come on math help the answer is 5sec^2x-4csc^2x? I have a calculus test coming up and I really would appreciate if someone could explain!

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Oh nvm I see my mistake!
 
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Hello, and welcome to MHB! (Wave)

The second answer you cited agrees with what W|A outputs as the derivative for the given function. If the two answers are equivalent, then the following will be an identity (I am assuming you mean something a bit different from the first answer):

$$-5\csc^2(x)=5\sec^2(x)-4\csc^2(x)$$

$$-\csc^2(x)=5\sec^2(x)$$

This is not an identity, so the first result you posted isn't correct. Where did this first result come from?
 

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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