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Continuity and Differentiability Test - 8

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Continuity and Differentiability Test - 8
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Weekly Quiz Competition
  • Question 1
    1 / -0.25

    The dervative of sin(x2 ) is

  • Question 2
    1 / -0.25

    If     where V = x2  - 2x + 3 then dy/dx is  

    Solution

    y = V3/3
    V = x2 - 2x + 3
    y = 1/3*v 3
    dy/dx = d/dx{1/3*(x2 -2x+3)3
    = 1/3*3*(x2 -2x+3)2 *(2x-2) 
    {d/dx(xn )=nx^n-1*dx/dx} 
    = 2(x-1)*(x2 -2x+3)2

  • Question 3
    1 / -0.25

    F(x) = tan (log x)
    F '(x) =

    Solution

    ∫tan(log x) dx
    log x = t
    x = et
    dx = et dt
    f(t) = ∫tan t dt
    f ’(t)= sec2 t
    = sec2 (log x)

  • Question 4
    1 / -0.25

    Which of the following function is not differentiable at x = 0?​

  • Question 5
    1 / -0.25

  • Question 6
    1 / -0.25

    y = log(sec + tan x)

    Solution

    y = log(secx + tanx)
    dy/dx = 1/(secx + tanx){(secxtanx) + sec2x }
    = secx(secx + tanx)/(secx + tanx)
    = secx

  • Question 7
    1 / -0.25

    If y=sin 3x . sin3  x then dy/dx is

    Solution

    Given: y = sin 3x . sin3

    = 3 sin2 x.sin 4x

  • Question 8
    1 / -0.25

    Solution

    LHD = limh →0 ((−h)2 sin(−1/h) −0)/−h
    limh →0 −h2 sin(1/h)/−h
    = limh →0 h ×sin(1/h)
    = 0
    RHD = limh →0 (h2 sin(1/h) −0)/h
    = limh →0 h ×sin(1/h)
    = 0
    RHD=LHD
    ∴f(x) is differentiable at x= 0

  • Question 9
    1 / -0.25

    ​Find the derivate of y = sin4 x + cos4 x

    Solution

    y=sin4x,  and z=cos4x
    So by using chain rule
    df(x)/dx = dsin4 x/dx + dcos4 x/dx
    =dy4 /dy * dy/dx + dz4 /dz * dzdx
    =dy4 /dy * dsinxdx + dz4 /dz * dcosx/dx
    =4y(4 −1) ⋅cosx+4z(4 −1)⋅(−sinx)
    =4sin3 xcosx −4cos3 xsinx
    =4sinxcosx(sin2 x −cos2 x)
    =2sin2x(−cos2x)
    =−2sin2xcos2x
    =−sin4x

  • Question 10
    1 / -0.25

    If   is equal to

    Solution

     Concept:

    Calculation:

    Hence, option (c) is correct.

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