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$f(x)=\left(x^{3}-1\right)^{3},$ find $(a) f^{\prime}(0)$ (b) $f^{\prime \prime}(0)$ (c) $f^{(3)}(0)$ (d) $f^{(4)}(0)$(e) $f^{(n)}(0)$ for $n>9$

(a) 0(b) 0(c) 18(d) 0(e) 0

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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question. Six states that f of X equals X to the third, minus one all to the third for part A. They would like you to find f prime of zero. So first taking that first derivative F prime of X is equal to three times X to the third minus one squared Ah, and then using the chain role multiplied by three x squared. Simplified. That's nine x squared times X to the third minus one squared Ah, and then taking f prime of zero. That would just be zero for Part B. They would like f double prime of zero. So taking the derivative again after Prime of X. Ah, we need to use the product rule here. So that comes out to nine X squared times two times x to the third minus one multiplied by three X squared plus next to the third minus one squared times 18 x. From there, we can start multiplying out to just simplify this so 54 x to the fourth times x to the third, minus one plus 18 x times x to the six minus two x to the third plus one again multiply everything out. So 54 x to the seventh, minus 54 x to the fourth plus 18 x to the seventh, minus 36 x to the fourth plus 18 x Then you have 72 x to the seventh, minus 90 x to the fourth plus 18 x and using that after all, prime of zero is just zero part C. They like, um, the third derivative. So after the third of X is equal to just take a derivative of this expression, we simplified. That's 504 x to the sixth, minus 3. 60 x to the third plus 18. Now third derivative of X or of zero, is now equal to 18. Taking the fourth derivative of X for part D, that would be five. I'm sorry. 3024 x to the fifth minus 1080 x to the fourth. Plugging in zero to our fourth derivative is just zero, then for part E when, after the end of zero. So the end derivative for n greater than nine, they would like you to find what that is. Essentially, what that would be is you're taking six more derivative. So after the fifth of X would have the highest power of X to the fourth F to the fourth back to the six of X be extra the third. Then, after the seventh of X would be X squared after the eighth of X would be X after the ninth of X would be when you get to a constant and then for after the 10 of X would just be zero, and anything after that would be zero. Therefore, after the end of zero for and greater than nine not equal to just greater, um is going to be just zero. And those are all your answers for Question six.

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