Practical:Product rule for differentiation - Revision history

Vipul: /* Criteria for computational mastery */

2023-12-05T17:53:04Z

Criteria for computational mastery

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	===Criteria for computational mastery===		===Criteria for computational mastery===

			<section begin="mastery"/>
	Since there's always some probability of computational errors, computational mastery doesn't mean always getting every computational problem correct. Moreover, your baseline rate of computational errors may differ from that of others, and we want to control for that baseline rate.		Since there's always some probability of computational errors, computational mastery doesn't mean always getting every computational problem correct. Moreover, your baseline rate of computational errors may differ from that of others, and we want to control for that baseline rate.

	One criterion for computational mastery is that differentiating products should be no more difficult (no more cognitively challenging and no more error-prone) than comparably lengthy differentiation problems involving sums of the same kinds of functions. This criterion has the benefit that it controls for both knowledge of the derivatives of individual functions and the baseline rate of computational errors.		One criterion for computational mastery is that differentiating products should be no more difficult (no more cognitively challenging and no more error-prone) than comparably lengthy differentiation problems involving sums of the same kinds of functions. This criterion has the benefit that it controls for both knowledge of the derivatives of individual functions and the baseline rate of computational errors.
			<section end="mastery"/>
	<section end="school"/>		<section end="school"/>

Vipul: /* Direct inline application of product rule for differentiation of three or more functions */

2023-12-05T17:51:20Z

Direct inline application of product rule for differentiation of three or more functions

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	The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.		The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.

			==Practice and mastery==

			===Self-generation of practice===

			While a lot of practice problems are available online and in textbooks, it's also possible to generate your own practice, essentially an infinite amount.

			* If you have access to a tool such as Mathematica or Wolfram Alpha that can do differentiation for you, you can take two arbitrary functions ''of different types'' that you know how to differentiate individually, then try to differentiate their product, and compare the answer with Mathematica or Wolfram Alpha. For instance, combine an algebraic function <math>x \mapsto x</math> with an exponential function <math>x \mapsto e^x</math>, to get <math>x \mapsto xe^x</math>. Try differentiating this yourself, then compare with the answer for [https://www.wolframalpha.com/input?i=differentiate+xe%5Ex differentiate xe^x] on Wolfram Alpha.
			* If you do not have access to a tool that can do differentiation for you, and you want to do practice you can verify, consider writing pairs of polynomial functions. Then try to differentiate their product in two ways: using the product rule, and expanding out the product and differentiating term-wise. Compare your answers.
			** You can also do similar practice when both functions are trigonometric functions related in a particular way; for instance, if they're both sine or cosine functions of linear functions. In this case, you can convert the product to a sum using trigonometric identities. You can then differentiate as a product and as a sum, and compare the answers you get.
			** You can also do similar practice when both functions are exponential functions related in a particular way; for instance, if they're exponential functions of polynomial functions, you can differentiate with the product rule or you can simplify by adding the exponents and then differentiate.

			===Criteria for computational mastery===

			Since there's always some probability of computational errors, computational mastery doesn't mean always getting every computational problem correct. Moreover, your baseline rate of computational errors may differ from that of others, and we want to control for that baseline rate.

			One criterion for computational mastery is that differentiating products should be no more difficult (no more cognitively challenging and no more error-prone) than comparably lengthy differentiation problems involving sums of the same kinds of functions. This criterion has the benefit that it controls for both knowledge of the derivatives of individual functions and the baseline rate of computational errors.

	<section end="school"/>		<section end="school"/>

Vipul at 17:39, 5 December 2023

2023-12-05T17:39:59Z

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	For instance, suppose we are given that <math>f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13</math>, we obtain that <math>(f \cdot g)'(1) = f'(1)g(1) + f(1)g'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109</math>.		For instance, suppose we are given that <math>f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13</math>, we obtain that <math>(f \cdot g)'(1) = f'(1)g(1) + f(1)g'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109</math>.
	~~<section end="school"/>~~

	==Dealing with products of three or more functions==		==Dealing with products of three or more functions==
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	The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.		The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.
			<section end="school"/>

Vipul: /* Direct inline application of product rule for differentiation of three or more functions */

2023-12-05T17:38:55Z

Direct inline application of product rule for differentiation of three or more functions

← Older revision		Revision as of 17:38, 5 December 2023
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	The inline procedure will now give:		The inline procedure will now give:

	<math>f'(x) = \frac{dx}{dx} e^x \ln x \sin x + x \frac{d(e^x)}{dx} \ln x \sin x + xe^x \frac{d(\ln x)}{dx} \sin x</math>		<math>f'(x) = \frac{dx}{dx} e^x \ln x \sin x + x \frac{d(e^x)}{dx} \ln x \sin x + xe^x \frac{d(\ln x)}{dx} \sin x + xe^x \ln x \frac{d(\sin x)}{dx}</math>

	This simplifies to:		This simplifies to:

	<math>f'(x) = ~~xe^x \ln x \frac{d(\sin x)}{dx} +~~ e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x</math>		<math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x</math>

	Further simplifying, we get:		Further simplifying, we get:

Vipul: /* Direct inline application of product rule for differentiation of three or more functions */

2023-12-05T17:38:30Z

Direct inline application of product rule for differentiation of three or more functions

← Older revision		Revision as of 17:38, 5 December 2023
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	This simplifies to:		This simplifies to:

	<math>f'(x) = xe^x \ln x \frac{d(\sin x)}{dx} = e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x</math>		<math>f'(x) = xe^x \ln x \frac{d(\sin x)}{dx} + e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x</math>

	Further simplifying, we get:		Further simplifying, we get:

Vipul: /* Direct inline application of product rule for differentiation of three or more functions */

2023-12-05T17:38:13Z

Direct inline application of product rule for differentiation of three or more functions

← Older revision		Revision as of 17:38, 5 December 2023
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	The inline procedure will now give:		The inline procedure will now give:

	<math>f'(x) = \frac{dx}{dx} e^x \ln x \sin x + x \frac{d(e^x)}{dx} \ln x \sin x + xe^x \frac{d(\ln x)}{dx} \sin x + xe^x \ln x \frac{d(\sin x)}{dx} = e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>		<math>f'(x) = \frac{dx}{dx} e^x \ln x \sin x + x \frac{d(e^x)}{dx} \ln x \sin x + xe^x \frac{d(\ln x)}{dx} \sin x</math>

			This simplifies to:

			<math>f'(x) = xe^x \ln x \frac{d(\sin x)}{dx} = e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x</math>

			Further simplifying, we get:

			<math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>

	The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.		The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.

Vipul: /* Dealing with products of three or more functions */

2023-12-05T17:37:33Z

Dealing with products of three or more functions

← Older revision		Revision as of 17:37, 5 December 2023
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	# '''Plug into the product rule formula''': We get <math>f'(x) = g'(x)h(x) + g(x)h'(x) = (e^x + xe^x)(\ln x \sin x) + (xe^x) \left( \frac{1}{x}\sin x + \ln x \cos x\right)</math>		# '''Plug into the product rule formula''': We get <math>f'(x) = g'(x)h(x) + g(x)h'(x) = (e^x + xe^x)(\ln x \sin x) + (xe^x) \left( \frac{1}{x}\sin x + \ln x \cos x\right)</math>
	# '''Simplify the expression thus obtained''': We get <math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>.		# '''Simplify the expression thus obtained''': We get <math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>.

			===Direct inline application of product rule for differentiation of three or more functions===

			The formula to remember is:

			<math>\! (f_1 \cdot f_2 \cdot \dots \cdot f_n)'(x) = f_1'(x)f_2(x) \dots f_n(x) + f_1(x)f_2'(x) \dots f_n(x) + \dots + f_1(x)f_2(x) \dots f_{n-1}(x)f_n'(x)</math>

			We can think of it as a sum of products, each product involving all the original factors, and with the differentiation hopping from left to right, with it starting on the left-most factor in the first term, and reaching the right-most factor in the last term.

			Let's look at the ''same'' example as above:

			{{quotation\|Differentiate the function <math>f(x) := xe^x \ln x \sin x</math> with respect to <math>x</math>}}

			The inline procedure will now give:

			<math>f'(x) = \frac{dx}{dx} e^x \ln x \sin x + x \frac{d(e^x)}{dx} \ln x \sin x + xe^x \frac{d(\ln x)}{dx} \sin x + xe^x \ln x \frac{d(\sin x)}{dx} = e^x \ln x \sin x + xe^x \ln x \sin x + xe^x \frac{1}{x} \sin x + xe^x \ln x \cos x = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>

			The advantage of the direct inline procedure is that it involves fewer steps and in particular doesn't involve doing side computations, thereby keeping the control flow on the main problem. The disadvantage is that it requires correct recall and application of a more complex formula. In certain grading or evaluation contexts, the direct inline procedure may also present the problem that the grader or reviewer may not be familiar with the product rule for differentiation for a product of multiple functions, or may not consider it a tool that can be legitimately applied directly to a problem.

Vipul: /* Most explicit procedure applied to a product of three or more functions */

2023-12-05T17:30:49Z

Most explicit procedure applied to a product of three or more functions

← Older revision		Revision as of 17:30, 5 December 2023
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	# '''Identify the two functions''': We'll set <math>g(x) := xe^x</math> and <math>h(x) := \ln x \sin x</math>.		# '''Identify the two functions''': We'll set <math>g(x) := xe^x</math> and <math>h(x) := \ln x \sin x</math>.
	# '''Calculate the derivatives''': Using the product rule, we get <math>g'(x) = e^x + xe^x</math>, and <math>h'(x) = \frac{1}{x} \~~sinx~~ + \ln x \cos x</math>.		# '''Calculate the derivatives''': Using the product rule, we get <math>g'(x) = e^x + xe^x</math>, and <math>h'(x) = \frac{1}{x} \sin x + \ln x \cos x</math>.
	# '''Plug into the product rule formula''': We get <math>f'(x) = g'(x)h(x) + g(x)h'(x) = (e^x + xe^x)(\ln x \sin x) + (xe^x) \left( \frac{1}{x}\sin x + \ln x \cos x\right)</math>		# '''Plug into the product rule formula''': We get <math>f'(x) = g'(x)h(x) + g(x)h'(x) = (e^x + xe^x)(\ln x \sin x) + (xe^x) \left( \frac{1}{x}\sin x + \ln x \cos x\right)</math>
	# '''Simplify the expression thus obtained''': We get <math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>.		# '''Simplify the expression thus obtained''': We get <math>f'(x) = e^x \ln x \sin x + xe^x \ln x \sin x + e^x \sin x + xe^x \ln x \cos x</math>.

Vipul: /* Applying the product rule for differentiation to compute values at a point */

2023-12-05T17:30:38Z

Applying the product rule for differentiation to compute values at a point

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 Knowledge of the values (in the sense of ''numerical values'') <math>\! f(x_0), g(x_0), f'(x_0), g'(x_0)</math> at a specific point <math>x_0</math> is sufficient to compute the value of <math>(f \cdot g)'(x_0)</math>.
-For instance, suppose we are given that <math>f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13</math>, we obtain that <math>(f \cdot g)'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109</math>.
+For instance, suppose we are given that <math>f(1) = 5, g(1) = 11, f'(1) = 4, g'(1) = 13</math>, we obtain that <math>(f \cdot g)'(1) = f'(1)g(1) + f(1)g'(1) = 4 \cdot 11 + 5 \cdot 13 = 44 + 65 = 109</math>.
 <section end="school"/>

Vipul at 05:41, 5 December 2023

2023-12-05T05:41:29Z

← Older revision		Revision as of 05:41, 5 December 2023
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	# '''Calculate the derivatives''' of <math>f</math> and <math>g</math> separately, on the side.		# '''Calculate the derivatives''' of <math>f</math> and <math>g</math> separately, on the side.
	# '''Plug into the product rule formula''' the expressions for the functions and their derivatives.		# '''Plug into the product rule formula''' the expressions for the functions and their derivatives.
	# '''Simplify the expression thus obtained''' (this is optional).		# '''Simplify the expression thus obtained''' (this is optional in general, though it may be required in some contexts).

	Here is an example of a differentiation problem where we use this explicit procedure:		Here is an example of a differentiation problem where we use this explicit procedure: