MHB How Do You Differentiate (1+sin²x)³?

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To differentiate (1 + sin²x)³, apply the chain rule. The derivative is calculated as 3(1 + sin²x)² multiplied by the derivative of the inner function, which is d/dx[1 + sin²x]. The inner derivative simplifies to 2sin(x)cos(x), leading to the final expression of 3(1 + sin²x)² * 2sin(x)cos(x). This method effectively demonstrates the application of the chain rule in calculus. Understanding these steps is crucial for mastering differentiation techniques.
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can someone go through the steps how to differentiate (1+sin²x)³
 
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markosheehan said:
can someone go through the steps how to differentiate (1+sin²x)³
It's multiple uses of the chain rule:
[math]\frac{d}{dx} [ (1 + sin^2(x) )^3 ] [/math]

[math]= 3 (1 + sin^2(x) )^2 \cdot \frac{d}{dx} [ 1 + sin^2(x) ] [/math]

Can you finish from here?

-Dan
 
yes thanks
 

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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