MHB Did I Find the Derivative of (5t^2+1)^4 Correctly?

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The discussion focuses on finding the derivative of the function (5t^2 + 1)^4. The user initially presents their derivative calculation, which involves applying the chain rule. Other participants confirm that the user's approach is correct, validating the steps taken in the differentiation process. The consensus is that the derivative has been found accurately. Overall, the thread emphasizes the proper application of the chain rule in calculus.
coolbeans33
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can you help me find the derivative of: (5t^2+1)4

this is what I did:

(5t^2)(ln 5)(2t) * 4(5t^2+1)3

did I do this right?
 
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Re: derivative with many exponents

Yep! Looks good to me. You took the derivative of the outermost part then applied the chain rule.
 
Re: derivative with many exponents

coolbeans33 said:
can you help me find the derivative of: (5t^2+1)4

this is what I did:

(5t^2)(ln 5)(2t) * 4(5t^2+1)3

did I do this right?

Yep.
 

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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