Continuity and Differentiability

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alok kumar singh

Contributor-Level 10

65. Kindly go through the solution

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4 months ago

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alok kumar singh

Contributor-Level 10

64. Let y = ex + ex2 + … + ex5.

dydx=ddx (ex + ex2 + ex3 + ex4 + ex5).

=dexdx+dex2dx+dex3dx+ddxex4+ddxex5.

=ex+ex2dx2dx+ex3dx3dx+ex4dx4dx+ex5dx5dx.

= ex + 2x ex2 + 3x2ex3 + 4x3ex4 + 5eex4

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alok kumar singh

Contributor-Level 10

63. Let y = log (cos ex).

dydx=ddlog (cos ex)

=1cosexddx (cos ex)

= - sinexcosexdex.dx

= -ex [tan ex]

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alok kumar singh

Contributor-Level 10

62. Kindly go through the solution

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alok kumar singh

Contributor-Level 10

61.  Kindly go through the solution

Let y = ex3

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alok kumar singh

Contributor-Level 10

60. Kindly go through the solution

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alok kumar singh

Contributor-Level 10

59. Let y = exsinx

dydx=sinxdexdxexddxsinxsin2x

=sinxexexcosxsin2x

=ex (sinxcosx)sin2x.

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alok kumar singh

Contributor-Level 10

58. Kindly go through the solution

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alok kumar singh

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57. Kindly go through the solution

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alok kumar singh

Contributor-Level 10

56. Kindly go through the solution

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