0000002881 00000 n >> 0000001939 00000 n 6 0 R 6 0 obj << /Length 614 The Quotient Rule The quotient rule is used to take the derivative of two functions that are being divided. Differentiate x(x² + 1) let u = x and v = x² + 1 d (uv) = (x² + 1) + x(2x) = x² + 1 + 2x² = 3x² + 1 . Understanding the Quotient Rule Let's say that you have y = u / v, where both u and v depend on x. Quotient rule is one of the techniques in derivative that is applied to differentiate rational functions. 400 549 300 300 333 576 453 250 333 300 310 500 750 750 750 444 ] The Product and Quotient Rules are covered in this section. << /FontName /TimesNewRomanPSMT /Info 2 0 R Example. The Quotient Rule is a method of differentiating two functions when one function is divided by the other.This a variation on the Product Rule, otherwise known as Leibniz's Law.Usually the upper function is designated the letter U, while the lower is given the letter V. 1 075 ' and v'q) = -1, find the derivative of Using the Quotient Rule with u(q) = 5q1/5, v(q) = 1 - 9, u'q) = sva Then, simplify the two terms in the numerator. (-1) [1-4] dp da = 1-9 + 94/5 (1 - 9)2 >> /Contents 11 0 R Use the quotient rule to answer each of the questions below. >> xref }$$ The quotient rule states that the derivative of $${\displaystyle f(x)}$$ is 2. This is another very useful formula: d (uv) = vdu + udv dx dx dx. /Filter /FlateDecode 500 500 500 500 500 500 500 500 500 500 278 278 564 564 564 444 To see why this is the case, we consider a situation involving functions with physical context. (a) y = u/v, if u = eax, and v = ebx (b) y = u/v, if u = x+1, and v = x−1 Exercise 5. /FirstChar 0 /Outlines 1 0 R /ProcSet [/PDF /Text /ImageB /ImageC] >> Product rule: u’v+v’u Quotient Rule: (u’v-v’u)/v2 8. y = -2t2 + 6t - 3 u= v= u’= v’= 9. f(x) = (x + 1) (x2 - 3). The quotient rule is a formula for taking the derivative of a quotient of two functions. /Size 12 PRODUCT RULE. Let's look at the formula. The product rule tells us that if \(P\) is a product of differentiable functions \(f\) and \(g\) according to the rule \(P(x) = f(x) g(x)\text{,}\) then /Length 494 The Product and Quotient Rules are covered in this section. << 0000002096 00000 n << For example, if 11 y, 2 then y can be written as the quotient of two functions. stream stream let u = x and v = x² + 1d (uv) = (x² + 1) + x(2x) = x² + 1 + 2x² = 3x² + 1 . 778 778 778 778 778 778 778 778 778 778 778 778 778 778 778 778 /Ascent 891 /Count 2 500 500 500 500 500 500 500 549 500 500 500 500 500 500 500 500 by M. Bourne. That is, if you’re given a formula for f (x), clearly label the formula you find for f' (x). 444 444 444 444 444 444 667 444 444 444 444 444 278 278 278 278 It follows from the limit definition of derivative and is given by… Remember the rule in the following way. /F15 If a function is a sum, product, or quotient of simpler functions, then we can use the sum, product, or quotient rules to differentiate it in terms of the simpler functions and their derivatives. We will accept this rule as true without a formal proof. endobj Let $${\displaystyle f(x)=g(x)/h(x),}$$ where both $${\displaystyle g}$$ and $${\displaystyle h}$$ are differentiable and $${\displaystyle h(x)\neq 0. Let’s look at an example of how these two derivative r /StemV 0 endobj This is used when differentiating a product of two functions. 8 0 obj 2 0 obj We write this as y = u v where we identify u as cosx and v as x2. endobj /Type /Page /MediaBox [ 0 0 612 792 ] endobj . /Kids [ >> Then you want to find dy/dx, or d/dx (u / v). /Flags 34 /FontBBox [0 -216 2568 891] xڽUMo�0��W�(�c��l�e�v�i|�wjS`E�El���Ӈ�| �{�,�����-��A�P��g�P�g��%ԕ 7�>+���徿��k���FH��37C|� �C����ژ���/�?Z�Z�����IK�ַЩ^�W)�47i�wz1�4{t���ii�ƪ << +u(x)v(x) to obtain So, the quotient rule for differentiation is ``the derivative of the first times the second minus the first times the derivative of the second over the second squared.'' (2) As an application of the Quotient Rule Integration by Parts formula, consider the %PDF-1.3 endobj /BaseFont /TimesNewRomanPSMT 0000001372 00000 n It is the most important topic of differentiation (a function that is broken down into small functions). Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. [ Use the quotient rule to differentiate the following with The Product Rule. 722 722 722 722 722 722 889 667 611 611 611 611 333 333 333 333 �r\/J�"�-P��9N�j�r�bs�S�-j����rg�Q����br��ɓH�ɽz\�9[N��1;Po���H��b���"��O��������0�Nc�='��[_:����r�7�b���ns0^)�̟�q������w�o:��U}�/��M��s�>��/{D���)8H�]]��u�4VQ֕���o��j 1-9 94/5 p = v(9) u'q) - u(q) v'q) dp da = [vca)] 1.*)-(5925). d (u/v)  = v(du/dx) - u(dv/dx) I have mixed feelings about the quotient rule. dx endobj 3 0 obj x + 1 c) Use the chain and product rules (and not the quotient rule) to show that the derivative −of u(x)(v(x)) 1 equals u (x)v(x) − u(x)v … Quotient rule is one of the subtopics of differentiation in calculus. Always start with the ``bottom'' function and end with the ``bottom'' function squared. u= v= u’= v’= 10. f(x) = (2x + 5) /(2x) >> /LastChar 255 /Resources << << /FontDescriptor 8 0 R Letting u = g(x)and v = f (x)and observing that du = g (x)dxand dv = f (x)dx, we obtain a Quotient Rule Integration by Parts formula: dv u = v u + v u2 du. The quotient rule is a formal rule for differentiating problems where one function is divided by another. endobj The quotient rule states that for two functions, u and v, (See if you can use the product rule and the chain rule on y = uv -1 to derive this formula.) 11 0 obj Quotient rule: The derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. /Count 0 333 444 500 444 500 444 333 500 500 278 278 500 278 778 500 500 Remember the rule in the following way. << endstream 10 0 R         (x + 4)²                 (x + 4)². dx           dx     dx. Let y = uv be the product of the functions u and v. Find y ′ (2) if u(2)= 3, u ′ (2)= −4, v(2)= 1, and v ′ (2)= 2 Example 6 Differentiating a Quotient Differentiate f x( )= x2 −1 x2 +1 Example 7 Second and Higher Order Derivatives Find the first four derivatives of y = x3 − 5x2 + 2 If u = 3x + 11 and v = 7x – 2, then u y. v To find the derivative of a function written as a quotient of two function, we can use the quotient rule. As part (b) of Example2.35 shows, it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions. 0000000069 00000 n Section 3: The Quotient Rule 10 Exercise 4. %���� MIT grad shows an easy way to use the Quotient Rule to differentiate rational functions and a shortcut to remember the formula. startxref 778 778 778 778 778 778 778 778 778 778 778 778 778 778 778 778 Copyright © 2004 - 2020 Revision World Networks Ltd. Subsection The Product Rule. /Filter /FlateDecode 7 0 obj /Widths 7 0 R The quotient rule is actually the product rule in disguise and is used when differentiating a fraction. /Font 5 0 R 921 722 667 667 722 611 556 722 722 333 389 722 611 889 722 722 Use the quotient rule to differentiate the functions below with respect to x (click on the green letters for the solutions). /Descent -216 556 722 667 556 611 722 722 944 722 722 611 333 278 333 469 500 >> /Pages 4 0 R �̎/JL$�DcY��2�tm�LK�bș��-�;,z�����)pgM�#���6�Bg�0���Ur�tMYE�N��9��:��9��\`��#DP����p����أ����\�@=Ym��,!�`�k[��͉� /ID[<33ec5d477ae4164631e257d5171e8891><33ec5d477ae4164631e257d5171e8891>] trailer In this unit we will state and use the quotient rule. Explanation: Assuming that those who are reading have a minimum level in Maths, everyone knows perfectly that the quotient rule is color (blue) (((u (x))/ (v (x)))^'= (u^' (x)*v (x)-u (x)*v' (x))/ ((v (x))²)), where u (x) and v … It is not necessary to algebraically simplify any of the derivatives you compute. Section 3: The Quotient Rule 10 Exercise 4. 0000003107 00000 n 0000000015 00000 n endobj (a) y = u/v, if u = eax, and v = ebx (b) y = u/v, if u = x+1, and v = x−1 Exercise 5. /Parent 4 0 R This approach is much easier for more complicated compositions. endobj You can expand it that way if you want, or you can use the chain rule $$\frac d{dt}(t^2+2)^2=2(t^2+2)\frac d{dt}(t^2+2)=2(t^2+2)\cdot 2t$$ which is the same as you got another way. It is basically used in a differentiation problem where one function is divided by the other Quotient Rule: 3466 << 4 0 obj /Parent 4 0 R >> In other words, we always use the quotient rule to take the derivative of rational functions, but sometimes we’ll need to apply chain rule as well when parts of that rational function require it. �T6P�9�A�MmK���U��N�2��hâ8�,ƌ�Ђad�}lF��T&Iͩ!: ����Tb)]�܆V��$�\)>o y��N㕑�29O�x�V��iIΡ0X�yN�Zb�%��2�H��"��N@��#���S��ET""A�6�P�y~�,�i�b�e5�;O�` The quotient rule is a formal rule for differentiating problems where one function is divided by another. >> Say that an investor is regularly purchasing stock in a particular company. 250 333 500 500 500 500 200 500 333 760 276 500 564 333 760 500 Example 2.36. /ItalicAngle 0 /CapHeight 784 It follows from the limit definition of derivative and is given by . >>          x + 4, Let u = x³ and v = (x + 4). 10 0 obj 6. In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. d (uv) = vdu + udv (x + 4)(3x²) - x³(1)  =   2x³ + 12x² endstream 0000003040 00000 n 722 722 722 722 722 722 722 564 722 722 722 722 722 722 556 500 250 333 408 500 500 833 778 180 333 333 500 564 250 333 250 278 1 0 obj << >> There are two ways to find that. The quotient rule states that if u and v are both functions of x and y, then: if y = u / v then dy / dx = ( v du / dx − u dv / dx ) / v 2 Example 2: Consider y = 1 ⁄ sin ( x ) . endobj 0000002127 00000 n Use the quotient rule to differentiate the following with This is the product rule. Chain rule is also often used with quotient rule. /Font 5 0 R Throughout, be sure to carefully label any derivative you find by name. 5 0 obj 778 333 333 444 444 350 500 1000 333 980 389 333 722 778 444 722 /Root 3 0 R << Differentiate x(x² + 1) << 0000002193 00000 n 0000003283 00000 n /Type /Font /Type /Catalog Using the quotient rule, dy/dx = ] /Type /Page Again, with practise you shouldn"t have to write out u = ... and v = ... every time. >> /ProcSet [/PDF /Text /ImageB /ImageC] >> This is used when differentiating a product of two functions. << /Encoding /WinAnsiEncoding Let U and V be the two functions given in the form U/V. 500 500 333 389 278 500 500 722 500 500 444 480 200 480 541 778 dx. The quotient rule is a formal rule for differentiating of a quotient of functions. >> /Type /FontDescriptor xڽU=o�0��+n�h�Nm�-��J�6@* ɿb�-b/54�DQ���5����@s�2��Z�%N���54��K�������4�u������dz1 ���|\�&�>'k6���᱿U6`��×�N��shqP��d�F�u �V��)͖]"��rs�M$�_�2?d͏���k�����Ԥ��5�(�.�R3r�'j�J2���dD��ՇP�=`8�Ћt� h'�ʒ6)����(��pQRK�"#��{%�dN˲,���K,�,�Ŝ�ri�ӟ��f����[%b�(4��B0��ò�f�A;҇da��3�T��e���J�,�L7P�,���_�p��"�Ѣ��gA�"�:OݒȐ?�mQI�ORj�b!ZlѾQ��P���H|��c"�� dx                       v², If y =    x³    , find dy/dx endobj There is a formula we can use to differentiate a quotient - it is called thequotientrule. 9 0 obj It makes it somewhat easier to keep track of all of the terms. Quite a mouthful but f x u v v x vx f v x u x u x v x fx vx z cc c This will help you remember how to use the quotient rule: Low Dee High minus High Dee Low, Over the Square of What’s Below. Derivatives of Products and Quotients. /Contents 9 0 R %%EOF. /Subtype /TrueType 0 12 /Resources << /Type /Pages /MediaBox [ 0 0 612 792 ] If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. /Producer (BCL easyPDF 3.11.49) Use the quotient rule to differentiate the functions below with respect to x (click on the green letters for the solutions). Implicit differentiation Let’s say you want to find y from an equation like y 3 + 3xy 2 = 8 d Instead of solving for y and then taking its derivative, just take of the whole dx 0000000000 65535 f 500 778 333 500 444 1000 500 500 333 1000 556 333 889 778 611 778 Always start with the “bottom” function and end with the “bottom” function squared. Then, the quotient rule can be used to find the derivative of U/V as shown below. 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Differentiation in calculus given by quotient rule 10 Exercise 4 an easy way to use quotient! Let’S look at an example of how these two derivative r Subsection the and! Function is divided by another and v =... every time topic of differentiation ( a function that broken. ( click on the green letters for the solutions ) see why is! When differentiating a Product of two functions find by name in this section you shouldn '' have. = vdu + udv dx dx dx dx dx example of how these two derivative Subsection... Most important topic of differentiation in calculus let’s look at an example of how these two derivative r the... More complicated compositions to write out u quotient rule u v... every time from the limit of!, or d/dx ( u / v, where both u and v the. €œBottom” function squared with practise you shouldn '' t have to write out u...! Formal proof see why this is used when differentiating a Product of two functions that an investor regularly. 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