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Grade 9 Exam  >  Grade 9 Tests  >  Test: Derivatives: Product And Quotient Rule - Grade 9 MCQ

Test: Derivatives: Product And Quotient Rule - Grade 9 MCQ


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10 Questions MCQ Test - Test: Derivatives: Product And Quotient Rule

Test: Derivatives: Product And Quotient Rule for Grade 9 2024 is part of Grade 9 preparation. The Test: Derivatives: Product And Quotient Rule questions and answers have been prepared according to the Grade 9 exam syllabus.The Test: Derivatives: Product And Quotient Rule MCQs are made for Grade 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Derivatives: Product And Quotient Rule below.
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Test: Derivatives: Product And Quotient Rule - Question 1
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The derivate of the function 

Detailed Solution for Test: Derivatives: Product And Quotient Rule - Question 1

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 

Detailed Solution for Test: Derivatives: Product And Quotient Rule - Question 2

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

Detailed Solution for Test: Derivatives: Product And Quotient Rule - Question 3

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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