MHB Differentiating tan2x: Steps & Solutions

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To differentiate tan(2x), apply the chain rule and the derivative of tangent. The derivative of tan(x) is sec^2(x), and using the chain rule, the derivative of tan(2x) becomes sec^2(2x) multiplied by the derivative of 2x, which is 2. Thus, the final result is 2sec^2(2x). For clarity, tan(2x) can also be expressed as sin(2x)/cos(2x) if using the quotient rule. Understanding these steps is essential for correctly finding the derivative.
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How do you differenciate tan2x and what are the steps?
 
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Hello and welcome to MHB, Trix! :D

There are two rules you want to use here:

i) $$\frac{d}{dx}\left(\tan(x)\right)=\sec^2(x)$$

ii) $$\frac{d}{dx}\left(f(u(x))\right)=\d{f}{u}\d{u}{x}$$

Can you apply these rules, and get the correct derivative?
 
Trix said:
How do you differenciate tan2x and what are the steps?

I'm going to assume that the OP doesn't already know the derivative of $\displaystyle \begin{align*} \tan{(x)} \end{align*}$, so you would need to write it as $\displaystyle \begin{align*} \frac{\sin{(2x)}}{\cos{(2x)}} \end{align*}$ and apply 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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