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3.2: Move those Chains

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Calculus, Chapter 2, Lesson 5.

Problem 1 – Derivative Using the Power Rule

Recall the Power Rule ddx(xn)=nxn1\begin{align*}\frac{d}{dx}(x^n)=n \cdot x^{n-1}\end{align*}.

1. Based on the Power Rule, what do you think the derivative of f(x)=(2x+1)2\begin{align*}f(x) = (2x + 1)^2\end{align*} is?

Graph the derivative of the function and your conjecture about the derivative. Go to the Y=\begin{align*}Y=\end{align*} Editor. In y1\begin{align*}y1\end{align*}, type (2x+1)2\begin{align*}(2x+1)^{\land}2\end{align*}. In y2\begin{align*}y2\end{align*}, type nDeriv(y1(x),x)\begin{align*}(y1(x),x)\end{align*}. To access the nDeriv command, go to the Math menu (2nd\begin{align*}2^{nd}\end{align*} [MATH]) and select B:Calculus > A:nDeriv(. In y3\begin{align*}y3\end{align*}, type your conjecture for the derivative of f(x)=(2x+1)2\begin{align*}f(x) = (2x + 1)^2\end{align*}. Highlight y1\begin{align*}y1\end{align*} and press F4\begin{align*}F4\end{align*} to unselect this function, and press []\begin{align*}[\blacklozenge]\end{align*} F3\begin{align*}F3\end{align*} to graph y2\begin{align*}y2\end{align*} and y3\begin{align*}y3\end{align*}. Note: The graph may take a minute to appear. If your conjecture is correct, the graphs of y2\begin{align*}y2\end{align*} and y3\begin{align*}y3\end{align*} will coincide. If your conjecture is incorrect, the graphs of y2\begin{align*}y2\end{align*} and y3\begin{align*}y3\end{align*} will not coincide.

2. Was your conjecture correct? If not, how can you change your conjecture to make it correct?

3. Expand the binomial (2x+1)2\begin{align*}(2x + 1)^2\end{align*}. Take the derivative of each term. How does this compare with your answer to Question 1?

Problem 2 – The Chain Rule

The following are ‘true’ statements that can be verified on the TI-89.

d((5x+7)3,x)=3(5x+7)25x\begin{align*}d((5x + 7)^\land 3,x)=3 \cdot(5x + 7)^\land 2 \cdot 5x\end{align*} true

d((x3+7)5,x)=5(x3+7)43x2\begin{align*}d((x^\land 3 + 7)^\land 5, x)=5 \cdot(x^\land 3 + 7)^\land 4 \cdot 3x^\land 2\end{align*} true

d((x2+6)4,x)=4(x2+6)32x\begin{align*}d((x^\land 2 + 6)^\land 4, x)=4 \cdot(x^\land 2 + 6)^\land 3 \cdot 2x\end{align*} true

4. What patterns do you see? Using any information that you can infer from these statements, create a rule for finding the derivative of these functions. Discuss the patterns you see and the rule you created with a partner.

5. Using your rule from Question 4, what is ddx((3x+2)2)\begin{align*}\frac{d}{dx} \left((3x+2)^2 \right)\end{align*}?

Verify your answer by typing your statement on the entry line of your TI-89. If you are correct, the TI-89 will return the word, ‘true’. If you are incorrect, the TI-89 will return a false statement. If you are incorrect, try again by editing your statement. You can copy your last command by selecting 2nd\begin{align*}2^{nd}\end{align*} ENTER.

6. What is ddx((7x+2)3)\begin{align*}\frac{d}{dx}((7x+2)^3)\end{align*}? Verify your answer.

7. What is ddx((4x2+2x+3)4)\begin{align*}\frac{d}{dx}((4x^2 + 2x + 3)^4)\end{align*}? Verify your answer.

The derivative rule you have just observed is called the Chain Rule. It is used to take the derivative of composite functions. The Chain Rule is ddx(f(g(x)))=f(g(x))g(x)\begin{align*}\frac{d}{dx}(f(g(x)))=f'(g(x)) \cdot g'(x)\end{align*}. First, take the derivative of the “outside function” at g(x)\begin{align*}g(x)\end{align*}. Then, multiply this by the derivative of the “inside function.”

Problem 3 – Homework Problems

1. ddx((4x3+1)2)=\begin{align*}\frac{d}{dx}((4x^3 + 1)^2)=\end{align*}
2. ddx((5x+10)7)=\begin{align*}\frac{d}{dx}((-5x + 10)^7)=\end{align*}
3. ddt((2t54t3+2t1)2)=\begin{align*}\frac{d}{dt}((2t^5 - 4t^3 + 2t -1)^2)=\end{align*}
4. ddx((x2+5)2)=\begin{align*}\frac{d}{dx}((x^2 + 5)^{-2})=\end{align*}
5. ddz((z33z2+4)3)=\begin{align*}\frac{d}{dz}((z^3 - 3z^2 + 4)^{-3})=\end{align*}

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