The Chain Rule. The Chain Rule. dy dy du dx du dx. For y = f (u) and u = g (x)

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1 AP Calculus Mrs. Jo Brooks The Chain Rule To find the derivative of more complicated functions, we use something called the chain rule. It can be confusing unless you keep yourself organized as you go through the problem, but once you get the process down it is easy. The Chain Rule dy dy du dx du dx For y = f (u) and u = g (x) In words, the Chain Rule states: if you have a function that consists of two or more embedded functions, to get the derivative you must take the product of the derivatives of each separate embedded function. We work from the outside in, taking the derivative of each "layer" function. That sounds complicated. Let me give a few examples that can explain it further. A. y = cos 5x This is the function u = 5x embedded in the function y = cos u outside function: y = cos u inside function: u = 5x The official rule: dy dy du dx du dx dy/dx = (-sin u) * (5) = -5 sin u dy/dx = -5 sin 5x y = cos 5x y = (- sin 5x) (5) y = -5 sin 5x deriv outside: dy/du = - sin u deriv inside: dx/du = 5 dy/dx = (deriv of outside)*(deriv of inside) Notice how the du's seem to cancel out to give you dy/dx! Take the product of each derivative. Simplify. Put the 5x back in for u. Don't forget this step. *** You can go directly to the answer if you want to just keep organized. Since we have the angle function embedded inside the cosine function, we will take the derivative of the cosine function times the derivative of the angle function. Note *** When you take the derivative of the outside function, the inside function remains unchanged.

2 AP Calculus Mrs. Jo Brooks B. y = sin 3 x y = (sin x) 3 Method One: y = 3 (sin x) (cos x) Method Two: dy dy du dx du dx This is the function u = sin x embedded in the power function y = u 3 We can also look at this function like this: y stuff 3 *** You can go directly to the answer if you want to just keep organized. This is the short method. Since we have the sine function embedded inside the power function, we will take the derivative of the power function times the derivative of the sine function. y 3stuff deriv of stuff Or you can do it more mathematically! This method takes a few more steps, but if it helps keep you organized, use it. The goal is to get the right derivative. This is the chain rule. We will take the derivative of the outside function * derivative of inside function outside function: y = u 3 deriv outside: dy/du = 3u inside function: u = sin x deriv inside: dx/du = cos x dy/dx = (3 u ) * (cos x) Take each derivative: dy/dx = (deriv outside) * (deriv inside) dy/dx = 3 sin x cos x Simplify and substitute back in for the u.

3 AP Calculus Mrs. Jo Brooks C. y = (x + ) -3 Method One: y 4 3 x x 4 y 6 Method Two: This is the function. We can also look at this function like this: y stuff 3 *** You can go directly to the answer if you want to just keep organized. Since we have the linear function embedded inside the power function, we will take the derivative of the power function times the derivative of the linear function. 4 y 3 stuff deriv of stuff Or you can do it more mathematically! dy dy du dx du dx This is the chain rule. We will take the derivative of the outside function * derivative of inside function outside: y = u -3 deriv outside: dy/du = -3u -4 inside: u = x+ deriv inside: dx/du = Take the product of the derivatives. dy/dx = (-3 u -4 ) * () dy/dx = -6 (x + ) -4 Simplify and substitute back in for the u. 3

4 AP Calculus Mrs. Jo Brooks D. y = cos (sin x) y = [- sin (sin x) ] (cos x) y = - cos x [sin (sin x) ] dy dx dy du * du dx dy/dx = (-sin u) * (cos x) dy/dx = - [sin (sin x)] cos x This is the function. We can also look at this function like this: y cos stuff *** You can go directly to the answer if you want to just keep organized. Since we have the sine function embedded inside the cosine function, we will take the derivative of the sine function times the derivative of the cosine function. y sin stuff deriv of stuff Or you can do it more mathematically! This is the chain rule. We will take the derivative of the outside function * derivative of inside function outside: y =cos u deriv outside: dy/du = - sin u inside: u = sin x deriv inside: dx/du = cos x Take the product of the derivatives. Simplify and substitute back in for the u. As you can see, some answers come out looking very strange. You will often have to use ( ) or [ ] to make your answer read correctly. It does not matter which method you use, as long as you take the time to get organized before you start and you understand how to come up with the correct derivative. I use the first method the trick is in being able to see the different embedded functions. Now take the time to go through the Flash Presentations on the Chain Rule. They will give you some quick practice on choosing the inner and outer functions. Once you have that mastered the chain rule is not difficult at all. When you finish those exercises, we will continue with the lesson on the chain rule. 4

5 AP Calculus Mrs. Jo Brooks The Chain Rule Continued Now that you have practiced, let's try a more complicated problem and take on three embedded functions. We will do these problems the short way by identifying the embedded functions and then going straight to the derivative. Just keep organized and you will be fine. Our "refined" Chain Rule is: Refined Chain Rule dy (deriv. of outside function) (deriv. of inside stuff) dx E. cosx y y stuff Stuff = + cos x Stuff = the angle of cosine y ' = (stuff)( deriv of stuff) y = ( + cos x) (- sin x) () y = - 4 ( + cos x) (sin x) This is our function. We can also look at the equation like this: y stuff Rewrite, at least mentally, the inside and outside functions. You need to know where the "stuff" is. In this case we have two layers of functions in the ( ) to identify too. Since we have the angle function embedded inside the cosine function which is embedded in the ( ) power function, we will take the derivative of the ( ) power function times the derivative of the inside of the ( ) times the derivative of the angle function. This one was a bit more complicated because you had three layers of functions: (stuff) Stuff = cos (stuff) Stuff = angle The derivatives of all three layers had to be multiplied together. As you can see, you get some strange answers and they can become complicated. My recommendation is to slowly look at the function and work your way from the outside in, until you cannot find any more functions to take the derivative of. Be sure to follow each individual differentiation rule (power, product, quotient, trig) completely and carefully. The trick on these is to be sure that you have identified all of the embedded functions so that you don't leave any of them out. F:

6 AP Calculus Mrs. Jo Brooks y = sin (7-5x) y = sin (stuff) y ' = cos (stuff) * (deriv of stuff) y = [cos (7 5x) ] (-5) Here is the function. Rewrite, at least mentally, the inside and outside functions. You need to know where the "stuff" is. In this case there is only one function in the ( ) to worry about. Since we have the angle function embedded inside the sine function, we will take the derivative sine function times the derivative of the angle function. y = -5 cos (7 5x) G: y = sec (3-8x) y = sec (stuff) y ' = sec (stuff) tan (stuff) * (deriv of stuff) y = [sec (3 8x) tan (3 8x)](-8) y = -8 sec (3 8x) tan (3-8x) This is our function. Rewrite, at least mentally, the inside and outside functions. You need to know where the "stuff" is. In this case there is only one function in the ( ) to worry about. Since we have the angle function embedded inside the secant function, we will take the derivative secant function times the derivative of the angle function. H: y 4 3x 9 y stuff x x 8 y 9 stuff deriv of stuff y y This is our function. Rewrite, at least mentally, the inside and outside functions. You need to know where the "stuff" is. In this case there is only one function in the ( ) to worry about. Since we have the function on the inside embedded inside the power function, we will take the derivative of the power function times the derivative of the inside function.

7 AP Calculus Mrs. Jo Brooks I: We have three substitutions (embedded functions) in this problem! t y cos5sin 3 y = cos(stuff) stuff = 5 sin (stuff) stuff = angle y = [ - sin (stuff)] [deriv of stuff] (deriv of stuff) t t y sin 5sin 5cos t t y sin 5sin cos This is the function. Rewrite, at least mentally, the inside and outside functions. You need to know where the "stuff" is. In this case we have both a sine and an angle function in stuff. Since we have the angle function embedded inside the sine function which is embedded in the cosine function, we will take the derivative of the cosine function times the derivative of the sine function times the derivative of the angle function. J. g (x) = 3 tan 4x g (x) = 3 tan (angle) x 3 sec angle deriv of angle 3sec4 4 sec 4 g g x x g x x Simplify. Here is our function. There is a constant out in front of the trig function, that we will just leave there. Identify the parts of the chain. Take the derivative of the tangent function times the derivative of the angle. These will get easier and eventually you will feel comfortable just skipping to the answer step. Keep organized and make sure you have identified all of the embedded functions, all of the "stuff"! 3

8 AP Calculus Mrs. Jo Brooks The General Power Rule Now that you are more comfortable with the chain rule, you will find that you will eventually be able to skip some steps and intuitively multiply by the derivatives you need. The General Power Rule is a generalization of what you can do when you have a set of ( ) raised to a power, (...) n. If the inside of the parentheses contains a function of x, then you have to use the chain rule. You will be able to get to the derivative by using the power rule with the (...) n and then also multiplying by the derivative of the inside of the parentheses. In other words, you are going to multiply by the old exponent and subtract one from the exponent over the ( ). Then you multiply by the derivative of the inside of the parentheses. dy n dx General Power Rule if y stuff n, then n stuff deriv of stuff Here is an example: gx34 9x 4 gx 3stuff 4 3 x 4 3stuff deriv of stuff g g x x Simplify. g x x Here is our function. There is a constant out in front of the parentheses that we will just leave there. The ( ) is being raised to the 4th power. The stuff inside of the ( ) is (4-9x). Find the derivative with the general power rule.

9 AP Calculus Mrs. Jo Brooks Often functions can be rewritten to take advantage of the general power rule. This function: Can be rewritten as: Now use the General Power Rule to find the derivative more easily: 3 f( x) x /3 x /3 f ( x) x x f( x) gt () 7 t f( x) x /3 3 g() t 7( ) t 3 () t 3 g() t 8t 3 3 gt () 7 3 What I did in the second one works best when there is a constant in the numerator. When you have a variable in the numerator, it really does not help that much and you may as well use the quotient rule. I will often refer to the general power rule, so you need to know what it is.

10 Can you find the embedded functions? Let s practice finding stuff! Click here to start.

11 For this function, how many embedded functions are there? y ln sin x none

12 Right there is only one embedded function here. The stuff is inside the parentheses. y y ln ln sin stuff Take the derivative using the chain rule: y stuff x * deriv of stuff * cos cos x y x cot x sin x sin x continue

13 Oops. There is one embedded function here. The stuff is inside the parentheses. y y ln ln sin stuff Take the derivative using the chain rule: x y stuff * deriv of stuff * cos cos x y x cot x sin x sin x continue

14 For this function, how many embedded functions are there? 4x g x e none

15 Right there is only one embedded function here. The stuff is in the exponent. Take the derivative using the chain rule: stuff * deriv of stuff g x e 4x g x e g x e stuff g x e * 4 4e 4 x 4 x continue

16 Oops. There is one embedded function here. The stuff is in the exponent. Take the derivative using the chain rule: stuff * deriv of stuff g x e 4x g x e g x e stuff g x e * 4 4e 4 x 4 x continue

17 For this function, how many embedded functions are there? y csc e 4x none

18 y csc Right there are two embedded functions here. One is the e function and the other is in the exponent. y y csc csc Take the derivative using the chain rule: e stuff 4x stuff stuff stuff stuff stuff cot st uff * e * deriv of stuff 4 x 4 x 4 x y csc e cot e * e * 4 y 4e csc e cot e 4 x 4x 4 x continue

19 y csc Oops! There are two embedded functions here. One is the e function and the other is in the exponent. y y csc csc Take the derivative using the chain rule: e stuff 4x stuff stuff stuff stuff stuff cot st uff * e * deriv of stuff 4 x 4 x 4 x y csc e cot e * e * 4 y 4e csc e cot e 4 x 4x 4 x continue

20 For this function, how many embedded functions are there? j 3sin cos none

21 Right there are two embedded functions here. One is the cosine function and the other is the angle. j 3sin cos 3sin stuff stuff j y 3cos Take the derivative using the chain rule: s * deriv of f deriv stuff tuff stuff stuf stuff * of y 3cos cos * -sin * 3 y cos cos sin continue

22 Oops. There are two embedded functions here. One is the cosine function and the other is the angle. j 3sin cos 3sin stuff stuff j y 3cos Take the derivative using the chain rule: s * deriv of f deriv stuff tuff stuff stuf stuff * of y 3cos cos * -sin * 3 y cos cos sin continue

23 For this function, how many embedded functions are there? h x f g x none

24 Right there is one embedded functions here. It is the function in the parentheses. Take the derivative using the chain rule: h x f g x stuff h x f stuff * deriv of stuff h x f * h x f g x g x Finish up!

25 Oops. There is one embedded functions here. It is the function in the parentheses. Take the derivative using the chain rule: h x f g x stuff h x f stuff * deriv of stuff h x f h x f g x g x * Finish Up!

26 Good job! Come back and review these from time to time. Identifying the embedded functions (i.e. finding the stuff) is an important skill. Review again if you want to

27 AP Calculus Mrs. Jo Brooks Homework Examples #, 5, 7, #: Find dy/dx for the function y sin(3x) y sin(3x ) This is our function. y sin( stuff ) I will rewrite it so that we can see the different layers. y cos(3x) 3 3cos(3x) Take the derivative: y ' = cos (stuff) * (derivative of stuff) #5: Find dy/dx for the function y 5cot x y 5cot y 5cot x stuff x x dy 5 csc dx dy 0x csc x dx y 5cot x This is our function. Since we will have to take the derivative of the angle, I will rewrite it with negative exponents. Now I will rewrite it so that we can see the different layers. Take the derivative. y ' = -csc (stuff) * (derivative of stuff) You do not have to rewrite the function to get rid of the negative exponents unless you want to. #7: Find dy/dx for the function y cos(sin x) y cos(sin x) y cos( stuff ) y sin(sin x) (cos x ) y cos x sin(sin x) This is our function. I will rewrite it so that we can see the different layers Take the derivative. y ' = -sin (stuff) * (derivative of stuff)

28 AP Calculus Mrs. Jo Brooks #: Find dy/dx for the function 5 3 y sin x cos x 5 3 y sin x cos x This is our function. y sin x cos x y stuff stuff 6 y 5sin x (cos x) 3cos x ( sin x) 6 y 5cos x sin x 3sin x cos 6 5cos sin 3sin cos x y x x x x First I will rewrite the function with different exponent notation to make it easier to see what is happening. Then rewrite to show our stuff. Take the derivative. y ' = -5 (stuff) -6 (deriv of stuff) - 3 (stuff) (deriv of stuff) You don't have to change the exponent notation back unless you want to.

29 AP Calculus Mrs. Jo Brooks Homework Examples #3, 7, 3 #3: Find dy/dx for the function 3 y sin xtan 4x 3 y sin xtan 4x This is our function. y x 3 sin tan 4x 3 y stuff tan stuff y x x 3 (sin ) sec (4 ) (4) (tan 4 x) 3 sin x (cos x) 3 4sin sec (4 ) 3cos sin tan 4x y x x x x First I will rewrite the function with different exponent notation to make it easier to see what is happening. Then rewrite to show our stuff. Take the derivative. We have to use the product rule here!! First = (stuff) 3 deriv of first = 3 (stuff) (deriv of stuff) second = tan (stuff) deriv of second = sec (stuff) (deriv of stuff) #7: Find dy/dx for the function y sin (3x ) y sin (3x ) sin(3x ) y This is our function. First I will rewrite the function with different exponent notation to make it easier to see what is happening. y sin( angle) y sin(3x) cos(3x ) 3 y 6sin(3x)cos(3x ) Then rewrite to show our stuff. Take the derivative. y ' = (sin (angle)) * (cos (angle) * (deriv of angle) #3: Find y '' for y = cot (3x - ). y = cot (3x - ) y csc 3x 3 y x x 3csc 3 3 csc 3 y 6csc 3x csc 3x cot 3x 3 y 8csc 3 cot 3 x x This is our function. Use the chain rule to get the first derivative. I rewrote the exponent so that you can see what we have to do. y ' = -csc (angle) * (deriv of angle) Take the derivative again. y '' = -6csc(angle) * (-csc (angle) cot (angle) * (deriv of angle)

30 AP Calculus Mrs. Jo Brooks Homework Examples #4, 45 #4: Find the equation of the line tangent to the curve defined by the parametric equations x = cos t and y = sin t at t = /4. x = cos t y = sin t dy cost dt These are our parametric equations. Find the derivative of each parametric equation with respect to t. dx sint dt dy cost dx sint cot t Use the chain rule to get dy/dx. At t = /4 m cot 4, Tangent line: y x Evaluate the derivative to find the slope of the tangent line and the point. Write the equation of the line.

31 AP Calculus Mrs. Jo Brooks #45: Find the equation of the line tangent to the curve defined by the parametric equations x = t and y t at t = /4. x = t y t dy t dt / These are our parametric equations. Find the derivative of each parametric equation with respect to t. dx dt dy t dx / At t = /4 m (), 4 Use the chain rule to get dy/dx. Evaluate the derivative to find the slope of the tangent line and the point. Write the equation of the line. Tangent line: y x 4

32 AP Calculus Mrs. Jo Brooks Homework Examples #5, 57, 63 #5: Let s cos 3 d. Evaluate ds/dt when and 5 dt s cos This is our equation. ds d Take the derivative with respect to time. sin dt dt ds 3 Put in the values we are given to find ds/dt. sin (5) 5 dt #57: The functions f and g and their derivatives have the following values at x = 0 and x = : x f (x) g (x) f ' (x) g '(x) / /3-8/3 Evaluate the following derivatives with respect to x: a) 5 f (x) - g (x) 5 f '(x) - g '(x) At x = 5 (-/3) - (-8/3) = f x g x 3 b) 3 f x3g xgx g x f x At x = 0 () [3() (/3)] + () 3 (5) = 6 c) f ( x) gx ( ) Here is the function. This is a straight forward derivative. Evaluate. Here is the function. To find the derivative we have to use the product rule and the chain rule for the g-function that had the power (derivative of the power * derivative of g(x)) Evaluate. Here is the function. gx ( ) gx ( ) f( x) f( xg ) '( x) At x = To find the derivative we have to use the quotient rule. Evaluate.

33 AP Calculus Mrs. Jo Brooks d) f (g (x)) f '(g (x)) * g '(x) at x = 0 f '() * g '(0) = -/3 * /3 = -/9 e) g (f (x)) g '(f (x)) * f '(x) at x = 0 g '() * f '(0) = -8/3 * 5 = -40/3 f) gx f x 3 g x f x g x f x At x = - [ ] -3 [-8/3 + -/3] = -6 f x g x g) f xg x g x at x = 0 f '(0 + g (0) ) [ + g '(0)] f '() [ + /3] = -/3 * 4/3 = -4/9 Here is the function. To find the derivative we have to use the chain. The outside is f (x) and the inside stuff is g (x). Evaluate. Here is the function. To find the derivative we have to use the chain. The outside is g (x) and the inside stuff is f (x). Evaluate. Here is the function. To find the derivative we have to use the chain rule for the (..)-function that has the power: (derivative of the power * derivative of inside stuff) Evaluate. Here is the function. To find the derivative we have to use the chain rule for the f-function that has the stuff inside the ( ): (derivative of the f * derivative of inside stuff) Evaluate.

34 AP Calculus Mrs. Jo Brooks #63: Suppose the velocity of a falling body is v k s m/sec (k a constant) at the instant the body had fallen s meters from its staring point. Show that the body's acceleration is constant. v k s / v ks dv a k s dt / a k s ks ds dt / / This is our velocity function. Rewrite it so that we will be able to take the derivative. To find acceleration, we need the derivative of velocity, but we need that derivative with respect to time, not position!!! Now, we know that ds/dt = v, so we can make that substitution. a k s k 0 Thus, a is constant. 3