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Calculus/second derivatives

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Expert: - 10/16/2009


Question
Find the first and second derivatives of the function.
(x^2+7)^(1/2)

1st derivative:  x(x^2 +7)^(-1/2)
2nd derivative:  -x^2 (x^2 +7)^(-3/2) apparently the second derivative is wrong. what is wrong with it??  

Answer
We have f(x) = (x²+7)^0.5.  This gives f'(x) = x(x²+7)^-0.5.

This is a function a(x) = x times a function b(x) = (x²+7)^-0.5.

This tells me that f"(x) = a'(x)b(x) + a(x)b'(x).

We can see that a'(x) = 1 and b'(x) = -x(x²+7)^-1.5.

Putting all of these together gives (x²+7)^-0.5 + x(-x)((x²+7)^-1.5).

The first number in the above equation is √(x²+7/√(x²+7),
and this needs to be (x²+7)√(x²+7)/(x²+7)².

The second number is -x²√(x²+7)/(x²+7)².

Since they have √(x²+7)/(x²+7)² in common,
we can say that adding them together gives (x²+7-x²)√(x²+7)/(x²+7)².

Note that the x² both cancel in front, giving 7√(x²+7)/(x²+7)².


This function is rationalized in denominator.
If the 3rd derivative was to be found, cancel the √(x²+7) in the top and bottom.

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