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$\operatorname{Find} \frac{d}{d x}\left(\frac{\left(3 x^{3}+12\right)^{2}\left(2 x^{5}-9\right)^{3}}{\sqrt{\frac{(x+1)^{5}}{\left(x^{2}+2\right)^{7}}}}\right)$ when $x=1$.

$$\frac{16689645 \sqrt{6}}{32}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

Missouri State University

Harvey Mudd College

University of Nottingham

Boston College

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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So what I see here is that I have two functions that I am dividing my top term in my bottom term. So when I'm dividing two different functions, I'm going to use the quotient rule here. And how am I going to apply the crucial role? Well, derivative of my top function is going to be Well, this has raised the one half power X squared minus one, raised the positive one half. So chain role first, bring down the outer function one half times the same function. 1/2 -1 is negative one half. Chain rule times the derivative the inside derivative of export is two X derivative, one is zero. So that whole term times the bottom function minus the top function Times the derivative of the bottom function which is just two x all divided by the quantity X squared plus one squared. Ok, so let's just simplify this a little bit. I know that one half and um to will cancel I'll also know that this will be written as X squared plus one. All over the square root of X squared minus one minus two X times the square root of x squared minus one. All divided by the quantity X squared plus one squared. Ok, given this, what am I doing now, I'm just plugging in the value X equals negative two. So replace each of these with negative to negative two square. This will be five, two squared negative, two squared negative times negative is positive. This will be the square 235 over the square to three minus um negative two times negative two is positive plus four times square +23. All divided by this will be five squared, which will be 25. So simplifying this a little bit multiplying the top. We'll explain this term by the score to three as well. Trying to simplify what I see here, this will cancel, this will be five plus four times the square 234 times three is going to be 12, all over 25 times square to three. So this will be 17/25 times square to three. And if I were to rationalize the denominator multiplied by the square 23 to the top and bottom 17 square 23 square 23 times square cancels all over 75. So this will be my value At X equals -2. So after I took the derivative, I just plugged in the value to figure out what it is.

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