Understanding the Derivative: Product Rule vs. Chain Rule

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SUMMARY

The discussion clarifies the simultaneous application of the product rule and the chain rule in calculus, specifically when differentiating the function sqrt(x^2 + 2). The derivative of sqrt(x^2 + 2) involves first applying the product rule and then the chain rule, resulting in the term 2x derived from the inner function x^2 + 2. The process is detailed through the substitution of u = x^2 and v = u + 2, leading to the derivative expression (1/2)v^(-1/2)v'. This highlights the necessity of understanding both rules for accurate differentiation.

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Where does this 2x (highlighted) come from? I thought this step was just the 'product rule', but it looks like the 'chain rule' was applied as (x^2 + 2)`. Were these two rules used simultaneously?

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First they used the product rule, and then they used the chain rule to find the derivative of sqrt(x^2+2). The 2x comes from the derivative of x^2+2.
 
That is actually a double application of the chain rule. To differentiate \sqrt{x^2+1}, let u= x2 and let v= u+ 1. \sqrt{x^2+ 1} becomes \sqrt{v}= v^{1/2}. It's derivative is (1/2)v-1/2v'. Of course, v'= u'= 2x.

(I am ignoring the x multipying the square root since you only asked about the chain rule.)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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