Find a and b such that the function  is both continuous and differentiable for all numbers in its domain. (By the way, what is the domain of g?)

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The domain of g is [0,). Any negative numbers will result in an imaginary output, since you would have to plug them into the square root. On to continuity. If the function is continuous, then its two pieces at x = 1 will have to meet; so, both of the two rules will have to have the same output at x = 1. So, plug in 1 for x and set them equal: So, a and b will be opposites. However, what are they? Well, we also know that g has to be differentiable at x = 1. Therefore, the function must not only be continuous there, it also must be smooth, since any type of sharp point or cusp will mean no derivative exists. Mathematically, this means that both the top and bottom rules must have the same derivative when x = 1. So, take the derivative of each, plug in 1 for x and set them equal: You know that b = 4 since a and b are opposites.

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Calculus II (BC): A Beguiling Derivative

Find  if .

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Solution:

The chain rule will be required. The "outer function" is arctangent. When you find the derivative of the "inner function", you have to use the product rule. Furthermore, finding the derivative of the natural log will require another application of the chain rule.

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