No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 0 answers
  • 1 answers

Yuvedha T 7 years, 5 months ago

Cosx^3.sin^2x^3 =Cosx^3.[d/dx(sin^2x^5)] + sin^2x^5[d/dx(cosx^3)] =cosx^3×2sinx^5×cosx^4×5x^4 + sin^2x^5×(-3x^2×sinx^3) = 10x^4. sinx^5.cosx^5.cosx^3-3x^2sinx^3.sin^2x^5. { formulaes used:₩ d/dx(cosx)=-sinx ₩d/dx(sinx)=cosx ₩d/dx(x^n=n.x^(n-1)}
  • 1 answers

Ankur Dahiya 7 years, 5 months ago

U can get all the answers from an ncert offline app
  • 0 answers
  • 0 answers
  • 1 answers

Rishabh Sahu 7 years, 5 months ago

Study all chapters ,practice questions from all topics , be sure to cover ncert miscellaneous and also spare some time for questions out of ncert from refreshers and cbse past papers.
  • 1 answers

Rishabh Sahu 7 years, 5 months ago

There are only students available here. There are no representatives of cbse here. If you want to know the date the best way is to contact regional cbse office.
  • 0 answers
  • 0 answers
  • 1 answers

Deepa Shree 7 years, 5 months ago

What is d query here
  • 1 answers

Deepa Shree 7 years, 5 months ago

If they are in ap then 2b/a+c=a/b+c+c/a+b Takinng lcm we get in rhs= 2a+b+c/(b+c)(a+b) now sbtract this we get
  • 0 answers
  • 0 answers
  • 1 answers

Amarjeet Singh 7 years, 5 months ago

Type the ques right
  • 2 answers

Kriti Chhetri 7 years, 5 months ago

|3A|=3^3|A| . Given |3A|=k|A| . ->> k|A|=3^3|A| . ->>k=3^3 . ->>k=27.

Tisha Arora 7 years, 5 months ago

27
  • 0 answers
  • 2 answers

Amarjeet Singh 7 years, 5 months ago

Those who have only two factors i.e 1 and the number itself

Anu Gupta 7 years, 5 months ago

2,3,5,7,11are prime no
  • 1 answers

Anu Gupta 7 years, 5 months ago

An binery operation in which produce as one element from two given element eg; addition , subtraction, multiplication,and division of number
  • 1 answers

Prabjeet Singh 7 years, 5 months ago

{tex}\text {Given equation is }y \text { log }x = x-y{/tex}            {tex}...(1){/tex}

{tex}\therefore x=y \text { log }x + y{/tex}

{tex}\Rightarrow x=y(\text {log }x+1)=y(1 + \text {log } x){/tex}          {tex}...(2){/tex}

{tex}\text {Differentiating Eqn. (1) using product rule, w.r.t. }x, \text { we get,}{/tex}

{tex}\cfrac {y}{x} + \text {log } x.\cfrac {dy}{dx} = 1-\cfrac {dy}{dx}{/tex}

{tex}\Rightarrow \text {log }x.\cfrac {dy}{dx} + \cfrac {dy}{dx} =1-\cfrac {y}{x}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx} \bigg(\text {log }x+1\bigg)=\cfrac {x-y}{x}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx}=\cfrac {x-y}{x(1+\text {log }x)}{/tex}

{tex}\text {From Eqn. (1) and Eqn. (2),}{/tex}

{tex}\Rightarrow \cfrac {dy}{dx} =\cfrac {y \text { log }x}{y(1+\text { log }x)(1+\text {log }x)}{/tex}

{tex}\Rightarrow \cfrac{dy}{dx}=\cfrac {\text { log }x}{(1+\text {log }x)^2}{/tex}{tex}\text {Proved}{/tex}

  • 0 answers
  • 3 answers

Devashish Agrawal 7 years, 5 months ago

True

Nitin Raghav 7 years, 5 months ago

True

Dhruv Bansal 7 years, 5 months ago

True
  • 0 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App