#$&* course mth 173 7/25 9 012. The Chain Rule
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Given Solution: `aIf t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx.. If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1. If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When you input these values in for this function you get the values of: Y(-2) = e^(-2^2) = e^4 = 54.598 Y(-1) = e^(-1^2) = e^1 = 2.718 Y(-.5) = e^(-.5^2) = e^.25 = 1.284 Y(0) = e^(0^2) = e^0 = 1 Y(.5) = e^(.5^2) = e^.25 = 1.284 Y(1) = e^(1^2) = e^1 = 2.718 Y(2) = e^(2^2) = e^4 = 54.598 All of these are the same values as the previous question. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aIf t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx. If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx. If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx. If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1. If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx. If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx. If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z. What would be the 'chain' of simple functions for the function y = cos ( ln(x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The “chain” of the simple function y = cos (ln(x)) would be: Z = ln(x) Y = cos (z) Since the variable isn’t, specified, I merely used the variable “z” just like in the previous question. Once you input the values of x in for the function z = ln(x) you then input those values in for the “z” for the function y = cos(z), thus giving you the same results as you would’ve gotten with y = cos (ln(x)). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z). Thus we have y = cos(z) = cos( ln(x) ). We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t introduce the concept of concept functions like in the given solution, but I do understand the concept. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The chain function for the functions y = (ln(t))^2 would be: Z = (ln(t)) Y = t^2 In this case, you get the same results as if you would’ve put the same values into the function y = (ln(t))^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2. Thus we have y = z^2 = (ln(t))^2. We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. What would be the chain of functions for y = ln ( cos(x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The chain of functions for this function would be: Z = cos(x) Y = ln(z) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z). Thus we have y = ln(z) = ln(cos(x)). This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): At first I was confused as to why things weren’t the other way around, then I realized that you have to focus on the variable itself and not the functions themselves. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. The rule for the derivative of a chain of functions is as follows: The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ). For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be (cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) . g(x) = x^2 so g'(x) = 2 x. f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2). Thus we obtain the derivative (cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) = 2 x * ( - sin ( x^2 ) ) = - 2 x sin ( x^2). Apply the rule to find the derivative of y = sin ( ln ( x ) ) . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this question, I used the first statement in the question as a model to follow by. The derivative of the “sin” portion of the function is cos and the derivative of the other portion of the function is 1/x. Following the model, I arrived at the final derivative of y’ = 1/x * cos(ln(x)). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWe see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ). Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x. Since f(z) = sin(z) we have f ' (z) = cos(z). Thus the derivative of y = sin( ln (x) ) is y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) = 1 / x * cos( ln(x) ). Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function. STUDENT QUESTION We have been given many different derivative functions in the last two assignments are these things that will always be given or should I make a chart of them and start learning them by memory. INSTRUCTOR RESPONSE There are not that many basic functions (power, exponential, logarithmic, sine, cosine cover most of it). You will need to know the derivatives of these functions and maybe a few more as we go along, and the rules for derivatives of product, quotient and composite functions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Find the derivative of y = ln ( 5 x^7 ) . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this function, f’ is 1/x and g’ is 35x^6. When put into the “chain rule” formula, you get a derivative of y’ = 35x^6 * (1/x)(5x^7). Since f’ is 1/g(x) the derivative can be simplified to y’ = 35x^6 * (1 / (5x^7)). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aFor this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus f ' (z) = 1 / z and g ' (x) = 35 x^6. We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7). So the derivative of y = ln( 5 x^7) is y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ]. Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Find the derivative of y = e ^ ( t ^ 2 ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In this function, f’ is e^ because e^ to any power is the same as its derivative. Also, g’ is 2t. Therefore, when you combine them to form the final derivative, you get y’ = 2t * e^ (t^2). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t. Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2). Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. Find the derivative of y = cos ( e^t ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given this function, the f’ is sin(x) and the g’ is still e^t, therefore the entire derivative is y’ = e^t * sin(e^t). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aWe have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t. Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t). Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function. STUDENT QUESTION: I am confused as to why the end result is - e^t sin (e^t). Is it because we just combined the multiplication?
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Given Solution: `aWe have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. Find the derivative of y = sin^4 ( x ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I broke this function down a little more than I did the others because it proved rather difficult to me. Each part of the function is as follows: F(x) = z^4 G(x) = sin(x) F’ = 4z^3 G’ = cos(x) Therefore, the derivative reads y’ = cos(x) * (4z^3)(sin(x)) but I remembered that if a sin or cos is raised to a power, that power then becomes a coefficient, therefore I simplified the derivative to y’ = 4 cos(x) * sin (x)^3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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Given Solution: `aThe composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power. We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x). STUDENT QUESTION: I am having a hard time figuring out the simplication of this one? Any help? INSTRUCTOR RESPONSE: You had the right function for the f and g functions, except that you reversed them: If f(z) = sin(z) and g(x) = x^4, then f(g(x)) = sin(x^4), and the derivative would indeed be 4 x^3 * cos(x^4). However the function here is sin^4(x), so f(z) = z^4 and g(x) = sin(x). This gives a very different result, as shown in the given solution. Regarding the simplification step cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x): a * b * c = b * a * c; this actually requires multiple applications of associative and commutative laws for addition, but it's something we're used to so we don't usually think about it that deeply. We know it is 'safe' to change the order of a sequence of multiplications. cos(x) * 4 ( sin(x) ) ^ 3 is just a sequence of multiplications. If a = cos(x), b = 4 and c = (sin(x))^3, then this expression is a * b * c. Then b * a * c = 4 cos(x) sin^3 (x). Note that sin^3(x) means (sin(x))^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. Find the derivative of y = cos ( 3x ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I partly based this solution on that of a previous question. When the cos(x) was in the front of that function, the sin(x) was then negative in the given derivative, so I thought that this one would be similar. F’ = -sin(x) and g’ = 3, therefore, the simplified derivative is y’ = -3 * sin(3x). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThis is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!