MHB What is the Derivative of xf(x) at x=4 Using the Product Rule?

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To find the derivative of xf(x) at x=4 using the product rule, apply the formula: (xf(x))' = f(x) + x f'(x). Given that f(4) = 7 and f'(4) = -2, substitute these values into the equation. This results in the derivative being calculated as 7 + 4(-2). Therefore, the derivative of xf(x) at x=4 is -1.
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I am having trouble getting started with this question.
Suppose that f(4)=7 and f′(4)=−2. Use the product rule to find the derivative of xf(x) when x=4. Thanks
 
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musad said:
I am having trouble getting started with this question.
Suppose that f(4)=7 and f′(4)=−2. Use the product rule to find the derivative of xf(x) when x=4. Thanks

It says exactly what to do, use the product rule on $\displaystyle \begin{align*} x\,f(x) \end{align*}$, you should get

$\displaystyle \begin{align*} \left[ x\,f(x) \right] ' &= x' \,f(x) + x\,f'(x) \\ &= 1\,f(x) + x\,f'(x) \\ &= f(x) + x\,f'(x) \end{align*}$

so what do you get when x = 4?
 

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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