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Test: Derivatives: Product And Quotient Rule - Commerce MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Derivatives: Product And Quotient Rule

Test: Derivatives: Product And Quotient Rule for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Derivatives: Product And Quotient Rule questions and answers have been prepared according to the Commerce exam syllabus.The Test: Derivatives: Product And Quotient Rule MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Derivatives: Product And Quotient Rule below.
Solutions of Test: Derivatives: Product And Quotient Rule questions in English are available as part of our Mathematics (Maths) Class 11 for Commerce & Test: Derivatives: Product And Quotient Rule solutions in Hindi for Mathematics (Maths) Class 11 course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt Test: Derivatives: Product And Quotient Rule | 10 questions in 10 minutes | Mock test for Commerce preparation | Free important questions MCQ to study Mathematics (Maths) Class 11 for Commerce Exam | Download free PDF with solutions
Test: Derivatives: Product And Quotient Rule - Question 1
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The derivate of the function 

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y = (1 + 1/x)/(1 - 1/x)
y = (x+1)/(x-1)
dy/dx = [(x-1)-(x+1)]/(x-1)2
= -2/(x-1)2

Test: Derivatives: Product And Quotient Rule - Question 2
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The derivate of the function 

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Given, y = (sinx+cosx)/(sinx−cosx)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)2 
[by quotient rule]
= [(−sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
− [(sinx−cosx)2 − (sinx+cosx)2]/(sinx−cosx)2
= − [(sin2x + cos2x − 2sinxcosx + sin2x + cos2x + 2sinxcosx)]/(sinx−cosx)2
= −2/(sinx−cosx)2
= -2/(sin2x + cos2x - 2sinxcosx)
= - 2/(1 - sin2x)

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Test: Derivatives: Product And Quotient Rule - Question 3
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Find the derivative of the given equation f(x) = x3 + x2 + 3

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using these two rulesdifferentiation of  f(x) = x3 + x2 + 3 will be 3x2 + 2x

Test: Derivatives: Product And Quotient Rule - Question 4
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The derivate of the function 
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Test: Derivatives: Product And Quotient Rule - Question 5
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Let (tan α) x + (sin α) y = α and (α cosec α) x + (cos α) y = 1 be two variable straight lines, α being the parameter. Let P be the point of intersection of the lines. In the limiting position when α→ 0, the coordinates of P are
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Test: Derivatives: Product And Quotient Rule - Question 6
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The derivative of the function sec2 x is:
Test: Derivatives: Product And Quotient Rule - Question 7
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Find w'(z) = 
Detailed Solution for Test: Derivatives: Product And Quotient Rule - Question 7

w’(z) = [(2-z)dy/dz(3z+9)- (3z+9)dy/dz(2-x)]/(2-z)2
[(2-z)(3) - (3z+9)(-1)]/(2-z)2
(6-3z+3z+9)/(2-z)2
= 15/(2-z)2

Test: Derivatives: Product And Quotient Rule - Question 8
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Test: Derivatives: Product And Quotient Rule - Question 9
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The derivate of  f(x) = 
Test: Derivatives: Product And Quotient Rule - Question 10
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The derivative of the function y = (px2+qx+r)/(ax+b) w.r.t. x is :
Detailed Solution for Test: Derivatives: Product And Quotient Rule - Question 10

y = (px2+qx+r)/(ax+b)
⇒ dy/dx = d/dx(px2+qx+r)/(ax+b)
= [(ax+b) d/dx(px2+qx+r) − (px2+qx+r) d/dx(ax+b)]/(ax+b)2
= [(ax+b)(2px+q) − (px2+qx+r).a]/(ax+b)2
= [(2apx2 + aqx + 2bpx + bq) − (apx2 + aqx + ar)]/(ax+b)2
= [apx2 + 2bpx + bq − ar]/(ax+b)2

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