University:
Rice University
Course:
MATH 212 | Multivariable Calculus
Academic year:

2023
Views:

113

Pages:

1
Author:

Keepiden

Higher Order Derivatives. Example Problem 2 Rewriting the square root that's raised to the 1/2 power. And then take the derivative, so f prime of x would be 1/2 x squared plus 8 to the negative 1/2, but we just learned that the inside's been changed.So this is chain rule. Again, the derivative of f of x can be determined by taking the second derivative of f of x and then plugging in 1. The first derivative is f prime of x = 1/2x2 + 8(-1/2). After changing the variables inside of the parenthesis, as we learned earlier, this becomes f prime of x = 1/2(x)(x)(-1/2) + 8. So let's take the derivative of "f" with respect to "x", which we'll call "f prime". The derivative of "f" with respect to x is negative 1/2 times "x" squared plus 8 over 3/2 times the derivative of the inside. So that's 2 times x plus 1 times the derivative of x. Now we need to determine where "f prime" equals 1. That'd be 1 times negative 1/2 and then 9 over 3/2, which is negative 3/2 divided by 2. And finally, we have 9 over 3/2 times 2 plus 9 divided by 2, which simplifies to -3 divided by 2. So negative 127 plus 1/3, which 1/3 is 9/27, so it looks like the answer here is 8/27. The nuance here was the fact that the first derivative resulted a chain, and that created a product when you took that second derivative.