MHB Derivative of 4/sqrt{x}: Step-by-Step Guide

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To find the derivative of the function 4/sqrt{x}, first rewrite it as 4x^(-1/2). Applying the power rule, the derivative is calculated by multiplying the exponent by the coefficient and decreasing the exponent by one. This results in the derivative being -2x^(-3/2). Therefore, the final expression for the derivative is -2/(2sqrt{x^3}). Understanding these steps is crucial for mastering derivatives involving fractional exponents.
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Steps for finding the derivative of 4/sqrt{x}
 
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schooler said:
Steps for finding the derivative of 4/sqrt{x}

Write it as $\displaystyle \begin{align*} 4x^{-\frac{1}{2}} \end{align*}$ and you should be able to continue :)
 

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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