Beginner having trouble taking derivative

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The discussion centers on difficulties in applying the chain rule and product rule together for derivatives. The user provides an example function g(x) and expresses confusion about combining these rules. It is clarified that the product rule should be applied first, followed by the chain rule for each derivative involved. The conversation emphasizes the importance of understanding how to sequentially apply these differentiation rules. Mastering this combination is crucial for solving complex derivative problems effectively.
Heinz21
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Hello, i am having trouble taking the derivative of functions with using the chain rule plus another rule such as the product rule... i know how to do chain rule by it self and product rule by itself but i am having trouble using them together.. here is an example
g(x)= ( 1+4x )5 ( 3+x-x2 )8
 
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Let f(x)=(1+4x)5 and h(x)=(3+x-x2)8

The g(x)=f(x)h(x) and g'(x)=f(x)h'(x) + f'(x)h(x)

Can you do the rest yourself?
 
What Mathman is saying is that you first must use The Product Rule for Differentiation.
Then each derivative [h ' (x) and f ' (x) ] will require using The Chain 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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