PRODUCT & QUOTIENT RULES CALCULUS 2. Dr Adrian Jannetta MIMA CMath FRAS INU0115/515 (MATHS 2) Product & quotient rules 1/13 Adrian Jannetta

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1 PRODUCT & QUOTIENT RULES CALCULUS 2 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Proct & quotient rules 1/13 Adrian Jannetta

2 Objectives In this presentation we ll continue learning how to differentiate more complicated functions. There are rules for differentiating procts of the form and quotients of the form y u(x) v(x) y u(x) v(x) After some simple examples we ll practice incorporating the chain rule into some more complicated procts and quotients. Proct & quotient rules 2/13 Adrian Jannetta

3 The proct rule To differentiate a function consisting of procts, like these: x sinx lnx cos x 3x 2 e x we must use a method called the proct rule. The Proct Rule Given a function of the form y uv where u and v are both functions of x,then the derivative is given by Or in a shorter form: u dv + v y uv + vu Proct & quotient rules 3/13 Adrian Jannetta

4 Proct rule derivation Consider the function yu(x)v(x) Ifδx is a small increase in x then there will be a corresponding increase in the values of u, v and y. y+(u+δu)(v+δv) Expanding the brackets on the RHS: y+ u(v+δv)+δu(v+δv) uv+uδv+vδu+δuδv If we subtract y from both sides (remembering that yuv) we get: u(v+δv)+δu(v+δv) y uv+uδv+vδu+δuδv uv uδv+vδu+δuδv Dividing both sides by δx we obtain: δx uδv+vδu+δuδv δx uδv δx + vδu δx +δuδv δx u δv δx + vδu δx +δu δv (1) δx If we letδx 0 then alsoδu 0,δv 0 and the third term on the RHS vanishes. The following small changes become exact: δv lim δx 0 δx dv, lim δx 0 Equation (1) becomes δu δx, lim δx 0 u dv + v δx Proct & quotient rules 4/13 Adrian Jannetta

5 Using the proct rule Use the proct rule to find the first derivative of the function y x 2 sinx. The RHS is composed of two functions. Let ux 2 and v sinx Differentiating each of these with respect to x we find: 2x and dv cosx The proct rule says u dv + v x2 cos x+2x sinx Proct & quotient rules 5/13 Adrian Jannetta

6 Proct rule Given f(x)4xe x, find f (x). Let Differentiate each to get: u4x and v e x 4 and dv ex The proct rule: f (x)u dv + v There are common factors here: f (x)4xe x + 4e x f (x)4e x (x+1) Proct & quotient rules 6/13 Adrian Jannetta

7 The quotient rule To differentiate a function consisting of quotients, like these: x+3 cos x e 2x 3lnx we can use a method called the quotient rule. The Quotient Rule Given a function of the form y u v sinx cos x where u and v are both functions of x,then the derivative is given by v u dv v 2 Or in a shorter form: y vu uv v 2 Proct & quotient rules 7/13 Adrian Jannetta

8 Quotient rule derivation Consider the function y u(x) v(x). Ifδx is a small increase in x then there will be a corresponding increase in the values of u, v and y. y+ u+δu v+δv Subtracting y from both sides (remembering that y u ) we get: v u+δu v+δv u v Combining the terms on the RHS: (u+δu)v u(v+δv) (v+δv)v uv+vδu uv uδv v 2 + vδv vδu uδv v 2 + vδv Dividing both sides by δx δx δx vδu uδv δx(v 2 + vδv) v δu δx uδu δx v 2 + vδv (2) If we letδx 0 thenδu 0,δv 0 and the 2nd term in the denominator vanishes. These small changes become exact: δv lim δx 0 δx dv, lim δx 0 Equation (2) becomes δu δx, lim δx 0 v u dv v 2 δx Proct & quotient rules 8/13 Adrian Jannetta

9 Using the quotient rule Find the derivative of the function y 3x+5 x The RHS is composed of two functions. Let Differentiating each with respect to x. The quotient rule says v u dv, so: v 2 u3x+5 and v x dv 3 and 2x (x 2 + 1)(3) (3x+5)(2x) (x 2 + 1) 2 3x x 2 10x (x 2 + 1) 2 3 3x 2 10x (x 2 + 1) 2 Proct & quotient rules 9/13 Adrian Jannetta

10 Differentiating tanx Find the derivative of ytanx by expressing it as a quotient We can use a trig identity to help in this case: Let Differentiating each with respect to x. y sin x cosx usin x and vcosx Using the quotient rule: dv cosx and sin x (cosx)(cosx) (sin x)( sinx) (cosx) 2 cos 2 x+sin 2 x cos 2 x But cos 2 x+sin 2 x1 so 1 cos 2 x sec2 x Proct & quotient rules 10/13 Adrian Jannetta

11 Incorporating the chain rule We might also need to apply the chain rule ring these problems. Chain rule within the proct rule Differentiate y(1 x 3 )e 2x Let u1 x 3 and v e 2x. To differentiate v we have to use the chain rule. 3x2 Now continue as usual: Taking a common factor of e 2x and dv 2e2x (1 x 3 )2e 2x + e 2x ( 3x 2 ) 2(1 x 3 )e 2x 3x 2 e 2x e 2x (2 3x 2 2x 3 ) Proct & quotient rules 11/13 Adrian Jannetta

12 Finding a tangent equation Find the equation of the tangent to the curve f(x)(x+4)e x at the point (0,4). Using the proct rule; let ux+4 and ve x Differentiate each of these with respect to x u 1 and v e x The first derivative f (x) uv + vu (x+4)( e x )+(e x )(1) e x (x+4)+e x f (x) e x (x+3) The gradient of the tangent at x 0 is f (0) 3. So, the equation of the tangent is At the point(0, 4): y 3x+c 4 3(0)+c c 4 This means the tangent equation at (0, 4) is y4 3x Proct & quotient rules 12/13 Adrian Jannetta

13 Test yourself... Now try to find for the following functions. 1 y x 3 lnx. 2 y 6x x y 6x 3x+8 4 Use quotient rule to find d [cot2x] (Hint: cotx cosx/sin x) Answers: 1 x2 (3lnx+1). 2 6(1 2x3 ) (x 3 + 1) (9x+16). 3x+8 4 2cosec 2 2x Proct & quotient rules 13/13 Adrian Jannetta