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  • 1 answers

Lipika Yadav 7 years, 5 months ago

1/2
  • 1 answers

Pooja Sharma 7 years, 5 months ago

1/x
  • 0 answers
  • 3 answers

Lipika Yadav 7 years, 5 months ago

Question is integrate under root tan x not tan x....

Kriti Chhetri 7 years, 5 months ago

log|sec x|

Lipika Yadav 7 years, 5 months ago

sec squarex/2under root tanx
  • 3 answers

Garima Singh 7 years, 5 months ago

4

Sonali Aggarwal 7 years, 5 months ago

4

Vishal Upadhyah 7 years, 5 months ago

4
  • 1 answers

Sambhav Jain 5 years, 8 months ago

The best way for this question is Don't do it by long division method 6238/28=222.78
  • 0 answers
A-
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  • 1 answers

Garima Singh 7 years, 5 months ago

Check on official site of cbse
  • 0 answers
  • 0 answers
  • 0 answers
  • 0 answers
  • 1 answers

Kriti Chhetri 7 years, 5 months ago

Root3/2
  • 1 answers

Saurabh Kumar 7 years, 5 months ago

U can get it from your playstore.u will get it in app and solution for free....i m also using it
  • 1 answers

Md Nishad 5 years, 8 months ago

Matrices is a represnt Data to each other Calculation
  • 2 answers

Harsh Deepak 7 years, 5 months ago

Pata nahi????

Ankit Kumae 7 years, 5 months ago

4
  • 3 answers

Abhinav Sharma 7 years, 5 months ago

-Pi÷6

Ankit Kumae 7 years, 5 months ago

-pie/6

Manan Bedi 7 years, 5 months ago

Pie /6
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  • 0 answers
  • 1 answers

Harish Nagar 7 years, 5 months ago

dy/dx =1/2(1-x/1+x)^1/2-1×d/dx(1-x/1+x) => dy/dx= 1/2underroot (1+x/1-x)×(1+x)d/dx(1-x)-(1-x)d/dx(1+x)/(1+x)^2 => dy/dx=1/2undroot(1+x/1-x)×(1+x)(-1)-(1-x)(1)/(1+x)^2 => dy/dx =1/2undroot(1+x/1-x)×-1-x-1+x/(1+x)^2 => dy/dx =-undroot (1+x/1-x)×1/(1+x)^2 => (1-x^2)dy/dx =-undroot (1+x)/(1-x)×1/(1+x)^2×(1-x^2) => (1-x^2)dy/dx =- undroot (1+x)/1-x) => (1+x^×)dy/dx =-y => (1-x^2)dy/dx +y=0 Hence proved Its very simple question ok
  • 1 answers

Manan Bedi 7 years, 5 months ago

Anything raised to the power 0 is 1
  • 0 answers
  • 1 answers

Manan Bedi 7 years, 5 months ago

D/dx(uv)=udv/dx+vdu/dx

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